### Stoichiometry - AP levelProblems #26-50

Problem #26: A 0.204 gram sample of a metal, M, reacts completely with sulfuric acid according to:

M + H2SO4 ---> MSO4 + H2

A volume of 213 mL of hydrogen is collected over water; the water level in the collecting vessel is the same as the outside level. Atmospheric pressure is 756.0 torr and the temperature is 25.0 °C. Calculate the molar mass of the metal.

The vapor pressure of water at various temperatures can be found in this table.

Solution:

1) Use Dalton's Law of Partial Pressures to determine the pressure of the dry H2:

756.0 − 23.8 = 732.2 torr

2) Converting that to atm gives:

732.2 torr / 760.0 torr/atm = 0.9634 atm

3) Use the ideal gas law:

PV = n RT

(0.9634 atm) (0.213 L) = (n) (0.08206 L atm / mol K) (298 K)

n = 0.0083915 mol H2

4) From the balanced equation, 1 mole of H2 is formed when 1 mol of metal M is reacted. Therefore, the moles of metal that reacted was:

0.0083915 mol M

5) The molar mass of M is:

0.204 g / 0.0083915 mol = 24.3 g/mol

The metal was magnesium.

Problem #27: A common way to obtain a pure metal from its impure metal oxide is to react the oxide with carbon, expressed generically as:

2MO(s) + C(s) ---> 2M(s) + CO2(g)

If 5.00 g of an unknown metal oxide (MO) reacted with excess carbon and formed 738 mL of CO2 at 200.0 °C and 0.978 atm, what is the identity of the metal?

Solution:

1) Using PV = nRT, solve for moles of CO2:

(0.978 atm) (0.738 L) = (n) (0.08206 L atm / mol K) (473 K)

n = 0.0186 mol

2) Determine moles of oxygen atoms in metal oxide:

For every one mole of CO2 produced, two moles of MO were consumed.

1 is to 2 as 0.0186 is to x

x = 0.0372 mol

3) Determine molar mass of MO:

5.00 g / 0.0372 mol = 134.44 mol

4) Determine atomic weight of M:

x + 16.00 = 134.44

x = 118.44 g/mol

M is tin.

Problem #28: A compound of P and F was analyzed as follows: heating 0.2324 g of the compound in a 378 cm3 flask turned all of it to gas, which had a pressure of 97.3 mmHg at 77 °C. Then, the gas was mixed with calcium chloride solution which turned all of the F to 0.2631 g of CaF2. Determine the molecular formula of the compound.

Solution:

1) Molecular weight of the gas:

(97.3 / 760) (0.378) = (n) (0.08206) (350)

n = 0.001684967 mol

0.2324 g / 0.001684967 mol = 138 g/mol

We will use this to move from the empirical formula to the molecular formula.

2) Mass of F:

(0.2631 g) (38.0 g / 78.074 g) = 0.128055 g

(38.0 g / 78.074 g) is called a gravimetric factor. It is the decimal percent of F in CaF2.

3) Moles of F:

0.128055 g / 19.0 g/mol = 0.00674 mol

4) Moles of P:

0.2324 g − 0.128055 g = 0.104345 g

0.104345 g / 31.0 g/mol = 0.003366 mol

5) Empirical formula:

P ---> 0.003366 / 0.003366 = 1
F ---> 0.00674 / 0.003366 = 2

PF2

6) Molecular formula:

PF2 weighs 31 + 38 = 69

138 / 69 = 2

PF2 times 2 = P2F4

Another P2F4 problem.

Problem #29: A metal chloride reacts with silver nitrate solution to give a precipitate of silver chloride according to following equation:

MCl2 + 2AgNO3 ---> M(NO3)2 + 2AgCl

When a solution containing 0.4750 g of metal chloride is made to react with silver nitrate, 1.435 grams of silver chloride are formed. Identify the metal.

Solution:

1) Determine moles of AgCl produced:

1.435 g AgCl / 143.32 g/mol = 1.001256 x 10-2 mol AgCl

2) Determine moles of MCl2 consumed:

1.001256 x 10-2 mol AgCl x (1 mol MCl2 / 2 mol AgCl) = 5.00628 x 10-3 mol MCl2

3) Determine molar mass of MCl2:

0.4750 g / 5.00628 x 10-3 mol = 94.88 g/mol

4) Determine the atomic weight and identity of M:

The atomic of M will be:

94.88 − (2 x 35.453) = 23.97 g/mol

M is magnesium

Problem #30: An unidentified metal M reacts with an unidentified halogen X to form a compound MX2. When heated the compound decomposes by the reaction:

2MX2 (s) ---> 2MX (s) + X2 (g)

When 1.12 g of MX2 is heated, 0.720 g of MX is obtained along with 56.0 mL of X2 gas (at STP).

a) What is the atomic mass and the identity of the halogen X?
b) What is the atomic mass and identity of the metal M?

Solution:

1) Calculate moles of X2 collected:

PV = nRT

(1.00 atm) (0.0560 L) = (n) (0.08206 L atm / mol K) (273 K)

n = 0.00250 mol

2) Calculate mass of X2 collected:

1.12 g − 0.720 g = 0.40 g

3) Calculate molecular weight of X2 and determine identity of X

0.40 g/0.00250 moles = 160 g/mol

half of 160 is 80, the atomic weight of X

bromine

4) Determine moles of MX:

MX : X2 molar ratio is 2 : 1

therefore, 0.0050 moles of MX

5) Calculate molar mass of MX:

0.720 g/0.0050 moles = 144 g/mole

6) Calculate atomic weight of M:

144 g/mol − 80 g/mol = 64 g/mole

copper

Problem #31: A metal sulfate has the formula M2SO4. 10.99 g of the compound was dissolved in water to make 500.0 cm3 of solution. A 25.0 cm3 sample was removed and reacted with an excess of BaCl2(aq) to produce a precipitate of BaSO4, which when dried had a mass of 1.167 g.

a) Determine the number of moles of BaSO4 precipitated.
b) Determine the concentration of M2SO4
c) Identify M

Solution:

1) Moles of BaSO4 precipitated from 25 cm3:

1.167 g / 233.4 g/mol = 0.005 mol

2) Moles of BaSO4 that would have precipitated from 500 cm3:

0.005 mol is to 25 as x is to 500

x = 0.100 mol of BaSO4

3) The molarity of the M2SO4 is this:

0.100 mol / 0.500 L = 0.200 M

The 0.100 mole comes from the fact that there is a 1:1 molar ratio between M2SO4 and BaSO4

M2SO4 + CaCl2 ---> BaSO4 + 2MCl

4) What is M?

0.1 mol of sulfate weighs 96.061 x 0.1 = 9.6061 g

10.99 − 9.6061 = 1.3839 g of M+ in solution

There were 0.2 mol of M+ in the 500 cm3 [from this: M2SO4(s) ---> 2M+(aq) + SO42-(aq)]

1.3839 g / 0.2 mol = 6.9 g/mol

Li2SO4

Problem #32: An element, X, forms two compounds with bromine: XBr2 and XBr4. When 10.00 grams of the XBr2 is reacted with excess bromine, 14.35 g of XBr4 is formed. Identify X.

Solution:

14.35 − 10.00 = 4.35 g of Br2 reacted

4.35 g / 159.808 g/mol = 0.02722 mol of Br2

XBr2 + Br2 ---> XBr4

XBr2 and Br2 react in a 1:1 molar ratio

Therefore, 10.00 g represents 0.02722 mol of XBr2

10.00 g / 0.02722 mol = 367 g/mol

In 367.377 g of XBr2, there is 159.808 g of bromine

367.377 − 159.808 = 207.569 <--- this is the atomic weight of X

Problem #33: Exactly 4.32 g of oxygen gas was required to completely combust a 2.16 g sample of a mixture of methanol and ethanol:

(1) How many moles of ethanol are contained within the sample?
(2) What is the percentage by weight of methanol in the sample?

Solution:

1) Balanced chemical equations:

CH3OH + 1.5O2 ---> CO2 + 2H2O
C2H5OH + 3O2 ---> 2CO2 + 3H2O

2) Let x = mass of methanol; let y = mass of ethanol

x + y = 2.16

[(1.5) (x/32)] + [(3) (y/46)] = 4.32/32

3) Using x = 2.16 − y:

[(1.5) ((2.16-y)/32)] + [(3) (y/46)] = 0.135

[(3.24 − 1.5y) / 32] + [3y / 46] = 0.135

4) Multiply each side by 32, then by 46:

(46) (3.24 − 1.5y) + (32) (3y) = 198.72

149.04 − 69y + 96y = 198.72

27y = 49.68

y = 1.84 g

x = 0.32 g

5) Answers to (1) and (2):

moles ethanol ---> 1.84 g / 46 g/mol = 0.040 mol

percent methanol ---> (0.32 g / 2.16 g) * 100 = 14.8%

Problem #34: A 3.41 g sample of a metallic element, M, reacts completely with 0.0158 mol of a gas, X2, to form 4.53 g MX. What are the identities of M and X?

Solution:

2M + X2 ---> 2MX

We know that M and X2 react in a 2:1 molar ratio.

Therefore, 0.0158 mol of X2 reacts with twice that many moles of M, 0.0316 mol

We can now determine the atomic weight of M:

3.41 g / 0.0316 mol = 107.9 g/mol

From the periodic table, we see that M is silver.

4.53 − 3.41 = 1.12 g <--- this is the mass of X in MX

We know that there is a 1:1 molar ratio between M and MX, therefore 0.0316 of MX was produced. Since we know the formula is MX, we know that 0.0316 mol of X is involved.

1.12 g / 0.0316 mol = 35.44 g/mol

This atomic weight is within experimental error for X to be chlorine.

Problem #35: When 2.3 moles of X reacts with 1.6 moles of Y, 71 grams of Z are produced. What is the molar mass of Z?

3X + 4Y ---> 5Z

This reaction has a 50% yield.

Solution:

1) Determine the limiting reagent:

X: 2.3 / 3 = 0.77
Y: 1.6 / 4 = 0.4

Y is the limiting reagent.

2) How much Z is produced at 100% yield:

4 is to 5 as 1.6 is to x

x = 2.0 mol

3) The reaction has only 50% yield, so 1.0 mole of Z was produced. The molar mass of Z is:

71 g / 1.0 mol = 71 g/mol

Problem #36: 20.0 mL of solution containing NaCl and KCl gave, on evaporation to dryness, 0.180 g of the mixed chlorides. 20.0 mL of the same solution gave 0.370 g of AgCl on treatment with a slight excess of the AgNO3 solution. Calculate, for the original solution, the mass per liter of both chlorides.

Solution:

1) Let x = mass NaCl and let y = mass KCl. from that, we get our first equation:

x + y = 0.180

2) Using AgCl, determine the moles of chloride ion in the solution:

moles AgCl ---> 0.370 g / 143.321 g/mol = 0.00258 mol

0.00258 mol gives the moles of chloride based on this:

Ag+(aq) + Cl¯(aq) ---> AgCl(s)
For every 1 mole of AgCl that precipitated, there was 1 mole of chloride in solution.

3) We can now write our second equation:

x / 58.4428 g/mol + y / 74.553 = 0.00258

x / 58.4428 ---> moles of NaCl
y / 74.553 ---> moles of KCl

Based on the 1:1 molar ratios between NaCl and Cl¯ (as well as KCl and Cl¯), we know that the two above divisions give the moles of chloride contributed from the NaCl and the KCl.

4) We will now substitute the first equation into the second and solve:

x = 0.180 − y

[(0.180 − y) / 58.4428] + [y / 74.553] = 0.00258

13.4 − 74.553y + 58.4428y = 11.2

16.1y = 2.2

y = 0.137 g <--- mass KCl in 20.0 mL
x = 0.180 − 0.137 = 0.0430 g <--- mass NaCl in 20.0 mL

mass KCl in 1 L ---> 0.137 g / 0.020 L = 6.85 g
mass NaCl in 1 L ---> 0.043 / 0.020 L = 2.15 g

Problem #37: You are given a mixture of three hydrated salts: Na2CO3 · 10H2O, MgSO4 · 7H2O, and CuSO4 · 5H2O. The total mass of the mixture is 12.123 grams. When the mixture is heated gently, the following two reactions occur:

Na2CO3 · 10H2O(s) ---> Na2CO3 · 7H2O(s) + 3H2O(g)
MgSO4 · 7H2O(s) ---> MgSO4 · H2O(s) + 6H2O(g)

After these reactions are complete, the mass of the mixture has decreased to 9.049 grams. This mixture is then heated more strongly, and the following additional reactions occur:

Na2CO3 · 7H2O(s) ---> Na2CO3(s) + 7H2O(g)
MgSO4 · H2O(s) ---> MgSO4(s) + H2O(g)
CuSO4 · 5H2O(s) ---> CuSO4(s) + 5H2O(g)

After this final heating, the mass of the mixture has decreased to 6.412 grams. From this information, calculate the masses of each of the three compounds in the original mixture.

Solution:

Comment: three simultaneous equations in three unknowns are required.

1) The first equation:

X + Y + Z = 12.123 g

where:

X = the mass of Na2CO3 · 10H2O
Y = the mass of MgSO4 · 7H2O
Z = the mass of CuSO4 · 5H2O

2) The second equation is developed based on the information gained from the first heating. Here is the second equation:

(232.0916 / 286.136) (X) + (138.3808 / 246.4696) (Y) + Z = 9.049 g

(232.0916 / 286.136) ---> this gravimetric factor is the decimal percent decrease in the original mass (the X) of the sodium carbonate decahydrate as it is changed to the heptahydrate. 232 is the molar mass of the heptahydrate and 286 is the molar mass of the decahydrate.

(138.3808 / 246.4696) ---> this gravimetric factor is the decimal percent decrease in the original mass (the Y) of magnesium heptahydrate as it is changed to the monohydrate. 246 is the molar mass of the heptahydrate and 138 is the molar mass of the monohydrate

There is no gravimetric factor for copper(II) sulfate pentahydrate since it did not lose any mass. It started with Z grams present and ended with Z grams present.

By the way, after the heating, the sample is now composed of Na2CO3 · 7H2O, MgSO4 · H2O, and CuSO4 · 5H2O.

3) The third equation is developed based on the information gained from the second heating. Here is the third equation:

(105.988 / 286.136) (X) + (120.366 / 246.4696) (Y) + (159.607 / 249.681) (Z) = 6.412 g

Notice that, in each gravimetric factor, the denominator is the molar mass of the original hydrate. This is because the 6.412 value references the entire loss of mass from the starting value of 12.123 g.

4) I'm going to rewrite the three equations, but I will use decimals rather than fractions:

X + Y + Z = 12.123

0.8111234X + 0.5614518Y + Z = 9.049

0.3704113X + 0.4883604Y + 0.6392437Z = 6.412

5) To calculate the answers, I used a 3 Equation System Solver. When I did that, I got the following answers (which I rounded off):

1.373 grams of Na2CO3 · 10H2O
6.418 grams of MgSO4 · 7H2O
4.332 grams of CuSO4 · 5H2O

6) I will start a step-by-step solution:

Rewrite the first equation:
Z = 12.123 − (X + Y)

Substitute into the second and third equations:

0.8111234X + 0.5614518Y + [12.123 − (X + Y)] = 9.049

0.3704113X + 0.4883604Y + 0.6392437[12.123 − (X + Y)] = 6.412

We now have two simultaneous equations in 2 unknowns.

And that is where I will leave it.

7) I must admit that I did not solve this problem on my own. When I found the problem in my notes, I did an Internet search which yielded some discussion about this problem. Look for links to 'www.chemicalforums.com'

Problem #38: A mixture of CuSO4 · 5H2O and MgSO4 · 7H2O is heated until all the water is lost. If 5.020 g of the mixture gives 2.988 g of the anhydrous salts, what is the percent by mass of CuSO4 · 5H2O in the mixture?

Solution:

1) For every mole of CuSO4 there are 5 moles of H2O, and for every mole of MgSO4 there are 7 moles of H2O. Therefore:

(5 * moles of CuSO4) + (7 * moles of MgSO4) = moles of H2O

2) Determine moles of water lost:

5.020 − 2.988 = 2.032 g

2.032 g / 18.015 g/mol = 0.112795 mol

3) Let X be the grams of anhydrous CuSO4. Therefore:

2.988 − X ---> the grams of anhydrous MgSO4

3) We can now substitute into the equation in step 1:

5 * (X / 159.6096) + 7 * [(2.988 − X) / 120.3686] = 0.112795

X / 159.6096 ---> moles of anhydrous CuSO4
(2.988 − X) / 120.3686 ---> moles of anhydrous MgSO4

4) Solve it:

(159.6096) (120.3686) * 5 * (X / 159.6096) + (159.6096) (120.3686) * 7 * [(2.988 − X) / 120.3686] = (0.112795) (159.6096) (120.3686)

(120.3686) * 5 * X + (159.6096) * 7 * (2.988 − X) = 2167.016

601.843X + 3338.3944 − 1117.2672X = 2167.016

515.4242X = 1171.3784

x = 2.27265 g of anhydrous CuSO4

5) Determine moles of anhydrous CuSO4:

2.27265 g / 159.6096 g/mol = 0.0142388 mol

6) Determine mass of water associated with 0.0142388 mol of anhydrous CuSO4:

(0.0142388 mol) (5) = 0.071194 mol of H2O

(0.071194 mol) (18.015 g/mol) = 1.28256 g

7) Determine mass of CuSO4 · 5H2O in the original sample and its percentage:

2.27265 g + 1.28256 g = 3.55521 g

(3.55521 / 5.020) * 100 = 70.82%

Problem #39: One mole of hydrocarbon Z is subjected to combustion. The product obtained is condensed and the resulting gaseous product occupied a volume of 89.6 L at STP. Oxygen required for this combustion was 145.6 L at STP. What is the molecular formula of Z?

Solution:

1) Let the formula for Z be this:

CxHy

2) Determine moles of CO2 and O2:

89.6 L / 22.4 L/mol = 4.00 mol CO2

145.6 L / 22.4 L/mol = 6.50 mol O2

3) The reaction thus far:

CxHy + 6.5O2 ---> 4CO2 + ___H2O

4) Determine the coefficient for H2O:

6.50 − 4.00 = 2.50 mol of O2 in the H2O

giving 5.00 mol H2O

5) Therefore:

CxHy + 6.5O2 ---> 4CO2 + 5H2O

x must be 4 and y must be 10

C4H10

Problem #40: A sample contains BaCl2 2H2O and NaCl. After 0.678 grams of the sample was heated, the residue weighed 0.648 grams. What is the percent of BaCl2 2H2O in the sample?

Solution:

1) Determine water driven off:

0.678 g − 0.648 g = 0.030 g H2O

2) Determine moles of water driven off:

0.030 g / 18.015 g / mol = 0.00166528 mol H2O

3) Determine moles of barium chloride dihydrate in sample:

(0.00166528 mol H2O) (1 mol BaCl2 2H2O / 2 mol H2O) = 0.00083264 mol BaCl2 2H2O

4) Determine grams of BaCl2 2H2O:

(0.00083264 mol) ( 244.2656 g / mol) = 0.2034 g

5) Determine percent BaCl2 2H2O in sample:

(0.2034 g / 0.678 g) * 100 = 30.0%

Problem #41: A 10.00 g sample of a mixture of CaCl2 and NaCl is treated with Na2CO3 to precipitate the CaCO3. This CaCO3 is heated to convert all the calcium to CaO and the final mass of CaO is 1.620 g. Determine the percent by mass of CaCl2 in the original mixture.

Discussion: We are given the mass of CaO, which will be used to calculate the mass of CaCO3 produced. Once we know the CaCO3 amount, we can determine the amount of CaCl2 present in the original mixture.

Solution:

1) When the mixture is heated, only this reaction occurs:

CaCO3 + heat ---> CaO + CO2

This balanced equation will be used as a math platform to show the needed calculation.

2) Over the CaO, enter 1.620 g and below the same, enter its gram-formula weight of 56.077 g/mol:

 1.620 CaCO3 ---> CaO + CO2 56.077

3) Over the CaCO3, enter X and below the same, enter its gram-formula weight of l00.086 g/mol:

 X 1.620 CaCO3 ---> CaO + CO2 l00.086 56.077

4) Cross multiply through the equation and solve for X:

 X 1.620 g ––––––––––– = ––––––––––– 100.086 g/mol 56.077 g/mol

X = 2.89137 g CaCO3 (I kept some guard digits)

5) We turn our attention to the equation that produces the calcium carbonate:

CaCl2 + Na2CO3 ---> CaCO3 + 2NaCl

6) Over the CaCO3, enter 2.89137 g and below the same, enter its gram-formula weight of l00.086 g/mol:

 2.89137 CaCl2 + Na2CO3 ---> CaCO3 + 2NaCl 100.086

7) Over the CaCl2, enter X and, below the same, enter its gram-formula weight of 110.984 g/mol

 X 2.89137 CaCl2 + Na2CO3 ---> CaCO3 + 2NaCl 110.984 100.086

8) Cross multiply through the equation and solve for X:

 X 2.89137 g ––––––––––– = ––––––––––– 110.984 g/mol 100.086 g/mol

X = 3.2062 g CaCl2 (I kept some guard digits)

9) Percent CaCl2 in original mixture:

(3.2062 g / l0.00 g) * 100 = 32.06% (o 4 sig figs)

Problem #42: How many milliliters of sodium metal, with a density of 0.971 g/mL, would be needed to produce 8.75 grams of hydrogen gas in a single replacement reaction with HCl?

Solution:

1) Write the balanced chemical equation:

Na + HCl ---> NaCl + 0.5H2 <--- note the 0.5 coefficient for the hydrogen gas

2) Determine moles of hydrogen gas produced:

 8.75 g ––––––––– = 4.340278 mol 2.016 g/mol

3) Use a Na to H2 mole ratio to determine moles of sodium that reacted:

 1 x –––– = ––––––––––– 0.5 4.340278 mol

x = 8.680556 mol Na consumed

4) Determine grams of Na in 8.680556 mol Na:

(8.680556 mol) (22.9898 g/mol) = 199.564 g

5) Determine volume of the sodium

 199.564 g ––––––––– = 205.524 mL 0.971g/mL

To three sig figs, the answer is 206 mL

Problem #43: A 5.75 g mixture contains both lithium fluoride, LiF, and potassium fluoride, KF. If the mixture contains 3.42 g fluorine, what is the mass of the KF in the mixture?

Solution:

1) Let "Z" be the mass (in grams) of the KF in the mixture.

2) Let (5.75 − Z) is the mass of LiF in the mixture.

3) Write expressions for the mass of F in each of the two compounds in the mixture.

 Z 1 mol KF 1 mol F 18.9984 g F ––– x ––––––––––– x ––––––– x ––––––––– = (0.327013Z) g F per g KF 1 58.0967 g KF 1 mol KF 1 mol F

 5.75 − Z 1 mol LiF 1 mol F 18.9984 g F ––––––– x ––––––––––– x ––––––– x ––––––––– = (4.21139 − 0.732415Z) g F per g LiF 1 25.9394 g LiF 1 mol LiF 1 mol F

4) Set the sum of the two expressions for grams of F equal to the given total grams of F (all units are in grams):

(0.327013Z) + (4.21139 − 0.732415Z) = 3.42

Z = 1.95 g KF

Problem #44: A mixture of two hydrocarbons, C8H18 (octane) and C7H8 (toluene), has a mass of 191 g. The hydrocarbon mixture is burned in excess oxygen to form a mixture of carbon dioxide and water that contains 1.56 times as many moles of carbon dioxide as water. Find the masses of C8H18 and C7H8 in the mixture.

Solution:

1) Let "Z" be the mass in grams of C8H18.

2) Let (191 − z) grams be the mass of C7H8.

3) Write the balanced chemical equation for the combustion of octane:

2C8H18 + 25O2 ---> 16CO2 + 18H2O

4) Write expressions for the molar amount of CO2 and H2O produced from combusting C8H18.

 Z 1 mol C8H18 16 mol CO2 ––– x –––––––––––––– x –––––––––– = (0.0700351Z) mol CO2 from C8H18 1 114.2285 g C8H18 2 mol C8H18

 Z 1 mol C8H18 18 mol H2O ––– x –––––––––––––– x –––––––––– = (0.0787894Z) mol H2O from C8H18 1 114.2285 g C8H18 2 mol C8H18

5) Write the balanced chemical equation for the combustion of C7H8.

C7H8 + 9O2 ---> 7CO2 + 4H2O

6) Write expressions for the molar amount of CO2 and H2O produced from combusting C7H8.

 191 − Z 1 mol C7H8 7 mol CO2 ––––––– x –––––––––––– x –––––––––– = (14.5108 − 0.075973Z) mol CO2 from C7H8 1 92.1384 g C7H8 1 mol C7H8

 191 − Z 1 mol C7H8 4 mol H2O ––––––– x –––––––––––– x –––––––––– = (8.29187 − 0.043413Z) mol H2O from C7H8 1 92.1384 g C7H8 1 mol C7H8

7) Add the two sources of CO2 and of H2O together:

total mol of CO2 ---> (0.0700351Z) + (14.5108 − 0.075973Z) = (14.5108 − 0.0059379Z)

total mol of H2O ---> (0.0787894Z) + (8.29187 − 0.043413Z) = (8.29187 + 0.0353764Z)

8) Write the expression for the ratio of total moles of CO2 to total moles of H2O, then set it equal to the ratio given in the problem statement:

 14.5108 − 0.0059379Z 1.56 ––––––––––––––––––– = –––– 8.29187 + 0.0353764Z 1

Z ≈ 25.8 g C8H18

(191 − Z) = 165 g C7H8 (to 3 sig figs)

Problem #45: 20 liters of SO2 reacts completely with 40 liters O2 to form SO3. At the end of the reaction, the combined volume of the substances is?

Solution:

1) Write the balanced chemical equation:

2SO2 + O2 ---> 2SO3

2) No information about temperature and pressure is provided. We, therefore, will assume that the volumes of the reactant (and products) are all at the same temperature and the same pressure.

3) The above allows us to invoke Avogadro's Hypothesis:

Equal volumes of gas, at equal temperatures and pressures, contain equal number of molecules

4) This means that the coefficients of a balanced chemical equation can be interpreted as volumes (a more general word than liters). Using the reaction in this problem:

two volumes of SO2 react with one volume of O2 to produce two volumes of SO3

5) The SO2 to O2 volume ratio is 2:1. Therefore:

20 L of SO2 react with 10 L of O2 (remember, 2:1 ratio)

The SO2 is completely consumed and 30 L of O2 remain.

6) The SO2 to SO3 ratio is 2:2. Therefore:

The 20 L of SO2 that reacted (and was completely consumed, making it the limiting reagent) will produce 20 L of SO3

7) 30 L of O2 left over and 20 litres of SO3 produced by the reaction means 50 L total.

Problem #46: How many NaBr formula units are formed when 50 NBr3 molecules and 57 NaOH formula units react? (Formula units are the equivalent of molecules for ionic compounds.)

2NBr3 + 3NaOH ---> N2 + 3NaBr + 3HBrO

Solution #1:

Of the two reactants, which is excess and which is the limiting one? The ratio is 2:3 and you have 50:57, so the 57 NaOH formula units is the limiting factor, it would use 2/3 of 57 = 38 NBr3 molecules.

Ratio of NaBr to NaOH is 3:3 or 1:1, so 57 NaOH formula units form 57 NaBr formula units.

Solution #2:

number of NaOH required = (3/2) * number of NBr3

(3/2) * 50 = 75

but we have just 57 NaOH. So, NaOH is limiting reagent

we will use NaOH in further calculation

from 1:1 ratio in reaction,

number of NaBr formed = number of NaOH = 57

Problem #47: According to the label on a bottle of concentrated hydrochloric acid, the contents are 36.0% by mass and has a density of 1.442 g/mL. What mass in kg of sodium hydrogen carbonate is needed to neutralize the spill if a bottle containing 5.127 L of HCl dropped and broke open.

Solution:

1) Balanced reaction (always a great place to start)

HCl + NaHCO3 ---> NaCl + H2O + CO2

Note the 1:1 molar ratio of HCl and NaHCO3

2) Big dimensional analysis set up (I will use 'sol' to indicate the HCl solution):

 5.127 L sol 1000 mL sol 1.422 g sol 36.0 g HCl 1 mol HCl 1 mol NaHCO3 84.007 g NaHCO3 1 kg ––––––– x ––––––– x ––––––– x ––––––– x ––––––– x ––––––– x ––––––– x ––––––– = 6.05 kg 1 1 L sol 1 mL sol 100 g sol 36.4609 g HCl 1 mol HCl 1 mol NaHCO3 1000 g

Problem #48: An ore containing magnetic, Fe3O4,was anlyzed by dissolving a 1.5419 g sample in concentrated HCI , giving a mixture of Fe2+ and Fe3+.

After adding HNO3 to oxidize any Fe2+ to Fe3+, the resulting solution was diluted with water and the Fe3+ precipitated as Fe(OH)3 by adding NH3. After filtering and rinsing, the residue was ignited , giving 0.8525 g of pure Fe2O3.

Calculate the % w/w Fe3O4 and % w/w Fe in the sample.

Solution:

1) All of the iron atoms in the final residue came from the original ore sample, regardless of all the intermediate reactions.

2) Calculate mass percent of Fe3O4:

 0.8525 g Fe2O3 1 mol Fe2O3 2 mol Fe 1 mol Fe3O4 231.53 g Fe3O4 1 –––––––––––– x –––––––––––– x –––––––––– x –––––––––– x –––––––––––– x ––––––– = 0.5346 = 53.46% Fe3O4 1 159.64 g Fe2O3 1 mol Fe2O3 3 mol Fe 1 mol Fe3O4 1.5419 g

3) Calculate mass percent of Fe:

 0.8525 g Fe2O3 1 mol Fe2O3 2 mol Fe 55.85 g Fe 1 –––––––––––– x –––––––––––– x –––––––––– x –––––––––– x –––––––––––– = 0.3868 = 38.68% Fe 1 159.64 g Fe2O3 1 mol Fe2O3 1 mol Fe 1.5419 g

Problem #49: Potassium thiosulfate, K2S2O3, is used to remove any excess chlorine from fibers and fabrics that have been bleached with that gas.

K2S2O3 + 4Cl2 + 5H2O ---> 2KHSO4 + 8HCl

(a) How many moles of K2S2O3 must react to produce 2.803 g of HCl?
(b) How many grams of Cl2 must react to produce 21.27 g of KHSO4?
(c) How many molecules of HCl are produced at the same time that 2.513 g of KHSO4 is produced?

Solution to (a):

 2.803 g HCl 1 mol HCl 1 mol K2S2O3 –––––––––– x ––––––––––– x ––––––––––– = 0.00961 mol K2S2O3 1 36.4609 g HCl 8 mol HCl

Solution to (b):

 21.27 g KHSO4 1 mol KHSO4 4 mol Cl2 70.906 g Cl2 ––––––––––––– x ––––––––––––––– x ––––––––––– x –––––––––– = 22.15 g Cl2 1 136.167 g KHSO4 2 mol KHSO4 1 mol Cl2

Solution to (c):

 2.513 g KHSO4 1 mol KHSO4 8 mol HCl 6.022 x 1023 molecules HCl ––––––––––––– x ––––––––––––– x –––––––––– x –––––––––––––––––––––– = 4.446 x 1022 molecules HCl 1 136.167 g KHSO4 2 mol KHSO4 1 mol HCl

Problem #50: Calculate minimum mass of P4S3 is required to produce at least 1.00 gm of each product.

P4S3 + 8O2 ---> P4O10 + 3SO2

Discussion:

This problem is solved by doing two mass-mass problems involving P4S3 in both and paired with each of the two products in two calculations. The greater amount of P4S3 will be the answer to the problem.

Solution:

1) How much P4S3 required to produce 1.00 g of P4O10?

 1.00 g P4O10 1 mol P4O10 1 mol P4S3 220.091 g P4S3 ––––––––––– x ––––––––––––– x –––––––––– x –––––––––––– = 0.775 g (to three sig figs) 1 283.886 g P4O10 1 mol P4O10 1 mol P4S3

2) How much P4S3 required to produce 1.00 g of SO2?

 1.00 g SO2 1 mol SO2 1 mol P4S3 220.091 g P4S3 ––––––––– x –––––––––– x –––––––––– x –––––––––––– = 1.145 g (to four sig figs) 1 64.063 g SO2 3 mol SO2 1 mol P4S3

3) To three sig figs, the answer to this problem is 1.14 g P4S3.