### Stoichiometry - AP levelProblems #11-25

Problem #11: Hydroxylammonium chloride reacts with iron(III) chloride, FeCl3, in solution to produce iron(II) chloride, HCl, H2O and a compound of nitrogen. It was found that 2.00 g of iron(III) chloride reacted in this way with 31.0 mL of 0.200 M hydroxylammonium chloride. Suggest a possible formula for the compound of nitrogen so produced.

Solution:

1) Determine moles of hydroxylammonium chloride (NH3OH+Cl¯) and iron(III) chloride:

(0.0310 L) (0.200 mol/L) = 0.0062 mol NH3OH+Cl¯

2.00 g / 162.204 g/mol = 0.01233 mol FeCl3

The key is to see that the moles of FeCl3 are double that of the hydroxylammonium chloride.

2) Determine the oxidation number of N in hydroxylammonium chloride:

N = -1

The determination of this is left to the reader.

By the way, we know that iron is reduced, so the nitrogen MUST be oxidized.

3) Allow the 0.0062 moles of nitrogen atoms to move from -1 oxidation state to zero:

this liberates 0.0062 mol of electrons, which go to reduce 0.0062 mol of Fe3+ ions (which is only half of the ions available)

4) Allow the 0.0062 mole of N atoms (because of step 3 just above, now at an oxidation state of zero) to move from zero to an oxidation state of +1:

this liberates another 0.0062 mol of electrons, which go to reduce 0.0062 mol of Fe3+ ions (which is the other half of the ions available)

5) We need nitrogen in the +1 oxidation state in our compound:

N2O

Problem #12: How many phosphate ions are in a sample of hydroxyapatite [Ca5(PO4)3OH] that contains 5.50 x 10-3 grams of oxygen?

Solution:

1) Determine moles of oxygen:

5.50 x 10-3 g divided by 16.00 g/mol = 3.4375 x 10-4 mol

2) Determine moles of hydroxyapatite:

the molar ratio between hydroxyapatite and oxygen is 1:13

3.4375 x 10-4 mol divided by 13 = 2.64423 x 10-5 mol of hydroxyapatite

3) Determine moles of phosphate ions:

the molar ratio between hydroxyapatite and phosphate is 1:3

2.64423 x 10-5 mol times 3 = 7.9326923 x 10-5 mol of phosphate ions

4) Determine number of ions:

7.9326923 x 10-5 mol times 6.022 x 1023 mol¯1 = 4.78 x 1019

Problem #13: A mixture consisting of only sodium chloride (NaCl) and potassium chloride (KCl) weighs 1.0000 g. When the mixture is dissolved in water and an excess of silver nitrate is added, all the chloride ions associated with the original mixture are precipitated as insoluble silver chloride (AgCl). The mass of the silver chloride is found to be 2.1476 g. Calculate the mass percentages of sodium chloride and potassium chloride in the original mixture.

Solution #1:

1) Set up this equation:

(x) (grams Cl from NaCl) + (1 − x) (grams Cl from KCl) = total grams chloride

x = grams of NaCl in original mixture
1 − x = grams of KCl in original mixture

grams Cl from NaCl = 35.5/58.4
grams Cl from KCl = 35.5/74.6
total grams chloride = (2.1476 g) (35.5/143.3)

The numbers in the denominators are the molar masses of NaCl, KCl and AgCl. The three ratios are called "gravimetric factors."

(x) (35.5/58.4) + (1 − x) (35.5/74.6) = (2.1476 g) (35.5/143.3)

2) Solve:

(x) (35.5/58.4) + (1 − x) (35.5/74.6) = (2.1476 g) (35.5/143.3)

0.6079x + 0.4759 − 0.4759x = 0.532

0.132x = 0.0561

x = 0.425 g

This is the mass of NaCl in the original mixture. This computes to 43% of the original mixture.

Solution #2:

1) Set up this equation:

mass of NaCl + mass of KCl = 1.000 g

x = grams of NaCl in original mixture
1 − x = grams of KCl in original mixture

Therefore:

(x) + (1 − x) = 1.000 g

2) Transform x and 1 − x as follows:

(x divided by 58.442 g/mol) times (143.321 g/mol) = 2.452x

((1 − x) divided by 74.551 g/mol) times (143.321 g/mol) = 1.992(1 − x)

Comment: using NaCl as an example, the transformation does this:

a) First, we calculate the moles of NaCl.
b) Since there is a 1:1 molar ratio between NCl and AgCl, this is also the number of moles of AgCl produced.
c) Multiply by the molar mass of AgCl to get the grams of AgCl produced from x grams of NaCl.

3) Write (then solve) this equation:

2.452x + 1.992(1 − x) = 2.1476

2.452x + 1.922 − 1.922x = 2.1476

0.53x = 0.2256

x = 0.426 g

Solution #3:

Graph the theoretical AgCl yield from one gram of 100% KCl through one gram of 100% NaCl with a few mixtures in-between to demonstrate linearity (or not) and interpolate your answer.

Comment: This would be fun to do on a spreadsheet someday.

Problem #14: Ammonia is produce industrially by reacting:

N2 + 3H2 ---> 2NH3

Assuming 100% yield, what mass of ammonia will be produced from a 1:1 molar ratio mixture in a reactor that has a volume of 8.75 x 103 L under a total pressure of 2.75 x 107 Pa at 455 °C.

Solution:

1) A 1:1 molar ratio means hydrogen is the limiting reagent. This is because a 1:3 ratio of nitrogen to hydrogen is required to fully react all the nitrogen.

2) Determine the initial pressure of hydrogen:

2.75 x 107 Pa = 2.75 x 104 kPa

2.75 x 104 kPa / 101.325 kPa/atm = 271.404 atm

271.404 atm / 2 = 135.702 atm

The divide by two is done because hydrogen makes up 50% of the reacting mixture.

3) Use PV = nRT:

(135.702 atm) (8.75 x 103 L) = (x) (0.08206) (728 K)

x = 19876.111 mol of hydrogen

4) Convert to amount of ammonia:

3:2 molar ratio for H2 : NH3

moles of NH3 = (19876.111 x 2) / 3 = 13250.74 mol

13250.74 mol x 17.0307 g/mol = 225669.4 g = 2.26 x 105 g

Problem #15: Upon heating, a 4.250 g sample loses 0.314 grams. Assuming the sample is BaCl2 · 2H2O and NaCl, calculate the mass percent of BaCl2 · 2H2O.

Solution:

1) Upon heating, only water is lost. Determine the moles of water lost:

0.314 g / 18.015 g/mol = 0.01743 mol of water

2) From the formula BaCl2 · 2H2O, we know:

2 moles of water per one mole of BaCl2

therefore 0.008715 mole of BaCl2

3) Determine grams, then percentage of barium chloride:

0.008715 mol x 244.2656 g/mol = 2.128775 g of BaCl2 · 2H2O

2.128775 / 4.250 = 50.09%

Problem #16: A 0.6118 g sample containing only MgCl2 and NaCl was analyzed by adding 145.0 mL of 0.1006 M AgNO3. The precipitate of AgCl(s) formed had a mass of 1.7272 g. Calculate the mass of each component (MgCl2 and NaCl) in the original sample.

Solution:

1) Using a gravimetric factor, determine the amount of chloride ion that preciptated:

1.7272 g times (35.453 / 143.321) = 0.42725366 g

2) Determine relative contribution of chloride by MgCl2 and NaCl:

for every three Cl¯ that react with Ag+:
two come from MgCl2
one comes from NaCl

Therefore:

magnesium chloride's contribution is 2/3
sodium chloride's contribution is 1/3

Please realize, this contribution is in terms of moles. So . . . .

3) Convert grams of chloride to moles:

0.42725366 g / 35.453 g/mol = 0.01205127 mol

4) Determine moles of NaCl in sample:

0.01205127 mol times one-third = 0.00401709 mol

5) Determine grams of NaCl in sample:

0.00401709 mol times 58.443 g/mol = 0.23477 g

to four sig figs: 0.2348 g

The mass of MgCl2 may be obtained by subtraction.

Problem #17: Ammonium nitrate and potassium chlorate both produce oxygen gas when decomposed by heating. Without doing detailed calculations, determine which of the two yields the greater

(a) number of moles of O2 per mole of solid and
(b) number of grams of O2 per gram of solid.

The unbalanced equations are:

NH4NO3(s) ---> N2(g) + O2(g) + H2O
KClO3(s) ---> KCl(s) + O2(g)

Solution:

1) Balance both equations:

2NH4NO3(s) ---> 2N2(g) + O2(g) + 4H2O
2KClO3(s) ---> 2KCl(s) + 3O2(g)

2) Write molar ratios:

NH4NO3 to O2 is 2:1
KClO3 to O2 is 2:3

3) Let the molar ratios be in terms of one mole of the solid:

NH4NO3 to O2 is 1:0.5
KClO3 to O2 is 1:1.5

In terms of moles, KClO3 produces more O2 than NH4NO3. In fact, KClO3 produces three times as much oxygen (compare 1.5 to 0.5).

5) Convert the moles of each molar ratio to grams:

NH4NO3 to O2 is 80.04 to 16.00
KClO3 to O2 is 122.55 to 48.0

6) Let the gram ratios be in terms of one gram of the substance:

NH4NO3 to O2 is 1 to 0.20
KClO3 to O2 is 1 to 0.39

In terms of grams, KClO3 produces oxygen approximately twice as fast (0.30 to 0.20) as NH4NO3.

Problem #18: An element X forms both a dichloride (XCl2) and a tetrachloride (XCl4), Treatment of 10.00 g XCl2 with excess chlorine forms 12.55 g XCl4. Calculate the atomic mass of X, and identify X.

Solution:

1) Write a balanced equation for the reaction:

XCl2 + Cl2 ---> XCl4

2) Determine grams, then moles of Cl2 that react:

12.55 g minus 10.00 g = 2.55 g

2.55 g / 70.906 g/mol = 0.035963 mol

3) Determine moles of XCl2 present:

Due to 1:1 molar ratio between XCl2 and Cl2, the moles of XCl2 equals 0.035963 mol

4) Determine the molecular weight of XCl2:

10.00 g / 0.035963 mol = 278.06 g/mol

5) Determine both atomic weight and identity of X:

278.06 g/mol minus 70.906 g/mol = 207.2 g/mol (rounded off to the 0.1 place)

Problem #19: Water is added to 4.267 g of UF6. The only products of the reaction are 3.730 g of a solid containg only uranium, oxygen, and fluorine and 0.970 g of a gas. The gas is 95.0% fluorine and the remainder is hydrogen.

a) What fraction of the fluorine of the orginal is in the solid and what fraction in the gas after the reaction?
b) What is the formula of the solid product?

Solution to a:

1) Calculate moles UF6 present:

4.267 g / 352.018 g/mol = 0.01212154 mol

2) Calculate grams of fluorine in UF6:

(0.01212154 mol) (113.988 g/mol) = 1.38171 g

The 113.988 comes from the fact that 6 F are in UF6

3) Calculate mass of fluorine in gas

(0.970 g) (0.950) = 0.9215 g

4) Calculate mass of fluorine in solid:

1.38171 g - 0.9215 g = 0.46021 g

5) Calculate percent fluorine in solid:

0.46021 g/ 1.38171 g = 33.307%

6) Calculate percent fluorine in gas:

100% - 33.307% = 66.693%

Solution to b:

1) Calculate mass of H2O reacted:

(3.730 g + 0.970 g) - 4.267 g = 0.433 g

2) Calculate mass of oxygen in solid product:

(0.433 g / 18.015 g/mol)(15.999 g/mol) = 0.38454438 g

3) Calculate mass of uranium in solid product:

3.730 g - (0.38454438 g + 0.46021 g) = 2.88524562 g

4) Calculate moles of U, F and O in solid product:

U: 2.88524562 g / 238.029 g/mol = 0.01212 mol

F: 0.4604 g/ 18.998 g/mol = 0.02423 mol

O: 0.38454438 g/15.999 g/mol = 0.02403 mol

5) To more clearly see the 1:2:2 ratio, simply divide by the smallest number:

U: 0.01212/0.01212 = 1
F: 0.02423/0.01212 = 1.999
O: 0.02403/0.01212 = 1.98

The formula of the unknown is UF2O2 and the overall reaction is:

UF6 + 2H2O --> UF2O2 + 4HF

Problem #20: A compound containing titanium and chlorine is analyzed by converting all the titanium into 1.20 g of titanium dioxide and all the chlorine into 6.45 g of AgCl. What is the simplest (empirical) formula for the original compound?

Solution:

By the way, note the use of millimoles rather than moles. Remember 1 mole equals 1000 millimoles.

1) This reaction happens:

TixCly ----> x TiO2 + y AgCl

2) Determine moles TiO2 formed:

1.20 g / 79.90 g mol¯1 = 15.02 mmol

3) Determine moles of AgCl formed:

6.45 g / 143.32 g mol¯1 = 45.00 mmol

4) Determine millimoles of Ti and Cl in original compound:

the Ti : TiO2 molar ratio is 1:1, therefore 15.02 mmol of Ti
The Cl : AgCl molar ratio is 1:1, therefore 45.00 mmol of Cl

5) The mole ratio of Ti to Cl in the compound is 15:45 or 1:3. Therefore:

the compound's formula is TiCl3

Problem #21: An unknown element X is found in two compounds, XCl2 and XBr2. In the following reaction:

XBr2 + Cl2 ---> XCl2 + Br2

when 1.5000 g XBr2 is used, 0.8897 g XCl2 is formed. Identify the element X.

Solution:

moles of XBr2 = moles of XCl2

1.500 / (x + 159.808) = 0.8897 / (x + 70.906)

(0.8897) (x + 159.808) = (1.500) (x + 70.906)

0.8897x + 142.1811776 = 1.5x + 106.359

0.6103x = 35.8221776

x = 58.70

Element X is Ni.

By the way, be careful. Take a look at Co and you'll see 58.93 and think that that is close enough. Nickel is 58.69. The Co/Ni pairing is one of three with the atomic weight goes down as you proceed from element to element. Ar/K and Te/I are the other two.

Problem #22: When the supply of oxygen is limited, iron metal reacts with oxygen to produce a mixture of FeO and Fe2O3. In a certain experiment, 20.00 g of iron metal was reacted with 11.20 g of oxygen gas. After the experiment the iron was totally consumed and 3.24 g oxygen gas remained. Calculate the amounts of FeO and Fe2O3 formed in this experiment.

Solution:

1) Determine moles of O2 that reacted:

11.20 g - 3.24 g = 7.96 g

7.96 g / 31.99886 g /mol = 0.248759 mol

2) There are two independent reactions occurring simultaneously:

2Fe + O2 ---> 2FeO
4Fe + 3O2 ---> 2Fe2O3

By the way, you might be tempted to write one equation:

3Fe + 2O2 --> FeO + Fe2O3

This would be incorrect as it masks the fact that the two oxygen to iron oxide molar ratios are different. Because the ratios are different, the calculation for FeO and Fe2O3 must be separate.

3) Let z be the mass fraction of Fe that produced FeO. Then 1-z is the mass fraction of Fe that produced Fe2O3 Determine the amount of oxygen consumed as (a) some Fe goes to FeO and (b) some Fe goes to Fe2O3:

(a) the FeO calculation:
 1 mol Fe 1 mol O2 (20.00 g Fe) (z) x –––––––––– x ––––––– = 0.179067z (mol O2 consumed by producing FeO) 55.8450 g Fe 2 mol Fe

(b) the Fe2O3 calculation

 1 mol Fe 3 mol O2 (20.00 g Fe) (1-z) x –––––––––– x ––––––– = (0.268601 - 0.268601z) (mol O2 consumed by producing Fe2O3) 55.8450 g Fe 4 mol Fe

Note that the amount of oxygen consumed is expressed in terms of the (unknown) mass fraction 'z.' The next step will determine the value of z.

4) Add the two amounts of O2 consumed and set it equal to the total moles of O2 reacted. Solve for 'z':

(0.179067z) + (0.268601 - 0.268601z) = 0.248759

-0.089534z = -0.019842

z = 0.221614

5) Determine (a) the amount of FeO produced and (b) the amount of Fe2O3 produced:

(a) the FeO calculation:
 1 mol Fe 2 mol FeO 71.8444 g Fe (20.00 g Fe) (0.221614) x –––––––––– x ––––––––– x ––––––––––– = 5.702 g FeO produced 55.8450 g Fe 2 mol Fe 1 mol Fe

(b) the Fe2O3 calculation:

 1 mol Fe 2 mol Fe2O3 159.6882 g Fe2O3 (20.00 g Fe) (0.778386) x –––––––––– x ––––––––– x ––––––––––– = 22.258 g Fe2O3 produced 55.8450 g Fe 4 mol Fe 1 mol Fe

6) Let's see if the Law of Conservation of Mass works:

20.00 g + 11.2 g - 3.24 g = 27.96 g
22.258 g + 5.702 g = 27.96 g

Yay!

7) Here's an incorrect solution to this problem.

Problem #23: A sheet of iron with a surface area of 525 cm2 is covered with a coating of rust that has an average thickness of 0.0021 cm. What minimum volume of an HCl solution, in mL, having a density of 1.07 g/mL and consisting of 14% HCl by mass is required to clean the surface of the metal by reacting with the rust? Assume that the rust is Fe2O3(s), that it has a density of 5.2 g/cm3, and that the reaction is:

Fe2O3(s) + 6HCl(aq) ---> 2FeCl3(aq) + 3H2O(ℓ)

Solution:

1) Volume of rust coating:

525 cm2 x 0.0021 cm = 1.1025 cm3

2) Mass of rust:

1.1025 cm3 x 5.2 g/cm3 = 5.733 g

3) Moles of rust:

5.733 g / 159.687 g/mol = 0.03590148 mol

4) Moles of HCl needed:

From the balanced equation, the Fe2O3 to HCl molar ratio is 1:6

1 is to 6 as 0.03590148 mol is to x

x = 0.21540888 mol

5) Mass of HCl needed:

0.21540888 mol x 36.4609 g/mol = 7.854 g

6) Mass of 14% solution required:

14 is to 100 as 7.854 is to x

x = 56.1 g

7) Volume of solution required:

56.1 g / 1.07 g/mL = 52.43 mL

Commentary: there are those teachers that absolutely insist on writing up a problem like the above in "dimensional analysis style." This means to string together all the calculations into one line. Here it is, done two different ways:

(525 cm2 x 0.0021 cm) (5.2 g/cm3) (1 mol / 159.687 g) (6 mol HCl / 1 mol Fe2O3) (36.4609 g / 1 mol) (100 / 14) (1 mL / 1.07 g)
 5.2 g 1 mol 6 36.4609 g 100 1 mL (525 cm2 x 0.0021 cm) ––––– x –––––– x –––– x ––––––––– x –––– x ––––– = 52.43 mL 1 cm3 159.687 g 1 1 mol 14 1.07 g

Make sure that the 159.687 gets used in a division. Also, note the 100/14. You have to multiply by 100, then divide by 14 before moving on.

There are those who insist that the DA style is clearer, making it pedagogically sounder to teach. The ChemTeam is not among that group.

Problem #24: A 1.42 g sample of a pure compound, with formula M2SO4, was dissolved in water and treated with an excess of aqueous calcium chloride, resulting in the precipitation of all the sulfate ions as calcium sulfate. The precipitate was collected, dried, and found to weigh 1.36 g. Determine the atomic mass of M. What element is it?

Solution:

The -2 charge on the sulfate ion is ignored.

1) Moles of sulfate precipitated:

(1.36 g CaSO4) / (136.1406 g CaSO4/mol) x (1 mol SO4 / 1 mol CaSO4) = 0.0099897 mol SO4

(1.36 g CaSO4) / (136.1406 g CaSO4/mol) ---> converts grams to moles
(1 mol SO4 / 1 mol CaSO4) ---> for every one mole of calcium sulfate produced, one mole of sulfate ion was used

2) Moles of M2SO4 holding that many moles of sulfate:

(0.0099897 mol SO4) x (1 mol M2SO4 / 1 mol SO4) = 0.0099897 mol M2SO4

3) Molar mass of M2SO4:

1.42 g / 0.0099897 mol = 142.1 g/mol

4) Molar mass of M:

SO4 = 96.0626 g/mol

142.1 g/mol M2SO4 - 96.0626 g/mol SO4 = 46.0 g/mol M2

(46.0 g/mol M2) / 2 = 23.0 g/mol M

sodium.

Problem #25: Calculate the volume change when iron is oxidized to Fe2O3 (d = 5.24 g/cm3). The density of Fe is 7.787 g/cm3.

Solution:

1) Write the balanced chemical equation for the reaction:

2Fe + 32O2 ---> Fe2O3

I decided to balance it with a fraction so as to use a Fe to Fe2O3 molar ratio of 2:1. If I had balanced the equation with whole numbers, the ratio I would have used would be 4:2. This would not affect the answer, my balancing choice was purely a stylistic one.

2) Let's start with 7.787 g (this is 1.00 cm3 of iron). Convert it to moles:

7.787 g / 55.845 g/mol = 0.13944 mol

3) Use 2:1 molar ratio:

 2 0.13944 mol ––– = –––––––––– 1 x

x = 0.06972 mol (of Fe2O3 produced)

4) Convert moles of Fe2O3 to grams:

(0.06972 mol) (159.687 g/mol) = 11.1334 g

5) Convert to cm3:

11.1334 g / 5.24 g/cm3 = 2.12 cm3

The volume changes from 1.00 cm3 to 2.12 cm3.