Problems #11-25

All Examples & Problems (no solutions)

**Problem #11:** Hydroxylammonium chloride reacts with iron(III) chloride, FeCl_{3}, in solution to produce iron(II) chloride, HCl, H_{2}O 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 (NH_{3}OH^{+}Cl¯) and iron(III) chloride:

(0.0310 L) (0.200 mol/L) = 0.0062 mol NH_{3}OH^{+}Cl¯2.00 g / 162.204 g/mol = 0.01233 mol FeCl

_{3}The key is to see that the moles of FeCl

_{3}are double that of the hydroxylammonium chloride.

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

N = -1The 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 Fe^{3+}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 Fe^{3+}ions (which is the other half of the ions available)

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

N_{2}O

**Problem #12:** How many phosphate ions are in a sample of hydroxyapatite [Ca_{5}(PO_{4})_{3}OH] 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 1:133.4375 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 1:32.64423 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 10^{23}mol¯^{1}= 4.78 x 10^{19}

**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 chloridex = grams of NaCl in original mixture

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

1 − x = grams of KCl in original mixtureTherefore:

(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) = 2.14762.452x + 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:

N_{2}+ 3H_{2}---> 2NH_{3}

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 10^{3} L under a total pressure of 2.75 x 10^{7} 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 10^{7}Pa = 2.75 x 10^{4}kPa2.75 x 10

^{4}kPa / 101.325 kPa/atm = 271.404 atm271.404 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 10^{3}L) = (x) (0.08206) (728 K)x = 19876.111 mol of hydrogen

4) Convert to amount of ammonia:

3:2 molar ratio for H_{2}: NH_{3}moles of NH

_{3}= (19876.111 x 2) / 3 = 13250.74 mol13250.74 mol x 17.0307 g/mol = 225669.4 g = 2.26 x 10

^{5}g

**Problem #15:** Upon heating, a 4.250 g sample loses 0.314 grams. Assuming the sample is BaCl_{2} **·** 2H_{2}O and NaCl, calculate the mass percent of BaCl_{2} **·** 2H_{2}O.

**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 BaCl_{2} **·** 2H_{2}O, we know:

2 moles of water per one mole of BaCl_{2}therefore 0.008715 mole of BaCl

_{2}

3) Determine grams, then percentage of barium chloride:

0.008715 mol x 244.2656 g/mol = 2.128775 g of BaCl_{2}·2H_{2}O2.128775 / 4.250 = 50.09%

**Problem #16:** A 0.6118 g sample containing only MgCl_{2} and NaCl was analyzed by adding 145.0 mL of 0.1006 M AgNO_{3}. The precipitate of AgCl(s) formed had a mass of 1.7272 g. Calculate the mass of each component (MgCl_{2} 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 MgCl_{2} and NaCl:

for every three Cl¯ that react with Ag^{+}:two come from MgCl_{2}

one comes from NaClTherefore:

magnesium chloride's contribution is 2/3

sodium chloride's contribution is 1/3Please 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 gto four sig figs: 0.2348 g

The mass of MgCl_{2} 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 O_{2}per mole of solid and

(b) number of grams of O_{2}per gram of solid.

The unbalanced equations are:

NH_{4}NO_{3}(s) ---> N_{2}(g) + O_{2}(g) + H_{2}O

KClO_{3}(s) ---> KCl(s) + O_{2}(g)

**Solution:**

1) Balance both equations:

2NH_{4}NO_{3}(s) ---> 2N_{2}(g) + O_{2}(g) + 4H_{2}O

2KClO_{3}(s) ---> 2KCl(s) + 3O_{2}(g)

2) Write molar ratios:

NH_{4}NO_{3}to O_{2}is 2:1

KClO_{3}to O_{2}is 2:3

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

NH_{4}NO_{3}to O_{2}is 1:0.5

KClO_{3}to O_{2}is 1:1.5

4) Answer to (a):

In terms of moles, KClO_{3}produces more O_{2}than NH_{4}NO_{3}. In fact, KClO_{3}produces three times as much oxygen (compare 1.5 to 0.5).

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

NH_{4}NO_{3}to O_{2}is 80.04 to 16.00

KClO_{3}to O_{2}is 122.55 to 48.0

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

NH_{4}NO_{3}to O_{2}is 1 to 0.20

KClO_{3}to O_{2}is 1 to 0.39

4) Answer to (b):

In terms of grams, KClO_{3}produces oxygen approximately twice as fast (0.30 to 0.20) as NH_{4}NO_{3}.

**Problem #18:** An element X forms both a dichloride (XCl_{2}) and a tetrachloride (XCl_{4}), Treatment of 10.00 g XCl_{2} with excess chlorine forms 12.55 g XCl_{4}. Calculate the atomic mass of X, and identify X.

**Solution:**

1) Write a balanced equation for the reaction:

XCl_{2}+ Cl_{2}---> XCl_{4}

2) Determine grams, then moles of Cl_{2} that react:

12.55 g minus 10.00 g = 2.55 g2.55 g / 70.906 g/mol = 0.035963 mol

3) Determine moles of XCl_{2} present:

Due to 1:1 molar ratio between XCl_{2}and Cl_{2}, the moles of XCl_{2}equals 0.035963 mol

4) Determine the molecular weight of XCl_{2}:

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)X is lead.

**Problem #19:** Water is added to 4.267 g of UF_{6}. 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 UF_{6} present:

4.267 g / 352.018 g/mol = 0.01212154 mol

2) Calculate grams of fluorine in UF_{6}:

(0.01212154 mol) (113.988 g/mol) = 1.38171 gThe 113.988 comes from the fact that 6 F are in UF

_{6}

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 H_{2}O 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 molF: 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 UF_{2}O_{2} and the overall reaction is:

UF_{6}+ 2H_{2}O --> UF_{2}O_{2}+ 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:

Ti_{x}Cl_{y}----> x TiO_{2}+ y AgCl

2) Determine moles TiO_{2} 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 : TiO_{2}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 TiCl_{3}

**Problem #21:** An unknown element X is found in two compounds, XCl_{2} and XBr_{2}. In the following reaction:

XBr_{2}+ Cl_{2}---> XCl_{2}+ Br_{2}

when 1.5000 g XBr_{2} is used, 0.8897 g XCl_{2} is formed. Identify the element X.

**Solution:**

moles of XBr_{2}= moles of XCl_{2}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 Fe_{2}O_{3}. 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 Fe_{2}O_{3} formed in this experiment.

**Solution:**

1) Determine moles of O_{2} that reacted:

11.20 g - 3.24 g = 7.96 g7.96 g / 31.99886 g /mol = 0.248759 mol

2) There are two independent reactions occurring simultaneously:

2Fe + O_{2}---> 2FeO

4Fe + 3O_{2}---> 2Fe_{2}O_{3}By the way, you might be tempted to write one equation:

3Fe + 2O

_{2}--> FeO + Fe_{2}O_{3}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 Fe

_{2}O_{3}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 Fe_{2}O_{3} Determine the amount of oxygen consumed as (a) some Fe goes to FeO and (b) some Fe goes to Fe_{2}O_{3}:

(a) the FeO calculation:

1 mol Fe 1 mol O _{2}(20.00 g Fe) (z) x –––––––––– x ––––––– = 0.179067z (mol O _{2}consumed by producing FeO)55.8450 g Fe 2 mol Fe (b) the Fe

_{2}O_{3}calculation

1 mol Fe 3 mol O _{2}(20.00 g Fe) (1-z) x –––––––––– x ––––––– = (0.268601 - 0.268601z) (mol O _{2}consumed by producing Fe_{2}O_{3})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 O_{2} consumed and set it equal to the total moles of O_{2} 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 Fe_{2}O_{3} 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 Fe

_{2}O_{3}calculation:

1 mol Fe 2 mol Fe _{2}O_{3}159.6882 g Fe _{2}O_{3}(20.00 g Fe) (0.778386) x –––––––––– x ––––––––– x ––––––––––– = 22.258 g Fe _{2}O_{3}produced55.8450 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 gYay!

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

**Problem #23:** A sheet of iron with a surface area of 525 cm^{2} 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 Fe_{2}O_{3}(s), that it has a density of 5.2 g/cm^{3}, and that the reaction is:

Fe_{2}O_{3}(s) + 6HCl(aq) ---> 2FeCl_{3}(aq) + 3H_{2}O(ℓ)

**Solution:**

1) Volume of rust coating:

525 cm^{2}x 0.0021 cm = 1.1025 cm^{3}

2) Mass of rust:

1.1025 cm^{3}x 5.2 g/cm^{3}= 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 Fe_{2}O_{3}to HCl molar ratio is 1:61 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 xx = 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 cm^{2}x 0.0021 cm) (5.2 g/cm^{3}) (1 mol / 159.687 g) (6 mol HCl / 1 mol Fe_{2}O_{3}) (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 cm ^{2}x 0.0021 cm)––––– x –––––– x –––– x ––––––––– x –––– x ––––– = 52.43 mL 1 cm ^{3}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 M_{2}SO_{4}, 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 CaSO_{4}) / (136.1406 g CaSO_{4}/mol) x (1 mol SO_{4}/ 1 mol CaSO_{4}) = 0.0099897 mol SO_{4}(1.36 g CaSO

_{4}) / (136.1406 g CaSO_{4}/mol) ---> converts grams to moles

(1 mol SO_{4}/ 1 mol CaSO_{4}) ---> for every one mole of calcium sulfate produced, one mole of sulfate ion was used

2) Moles of M_{2}SO_{4} holding that many moles of sulfate:

(0.0099897 mol SO_{4}) x (1 mol M_{2}SO_{4}/ 1 mol SO_{4}) = 0.0099897 mol M_{2}SO_{4}

3) Molar mass of M_{2}SO_{4}:

1.42 g / 0.0099897 mol = 142.1 g/mol

4) Molar mass of M:

SO_{4}= 96.0626 g/mol142.1 g/mol M

_{2}SO_{4}- 96.0626 g/mol SO_{4}= 46.0 g/mol M_{2}(46.0 g/mol M

_{2}) / 2 = 23.0 g/mol Msodium.

**Problem #25:** Calculate the volume change when iron is oxidized to Fe_{2}O_{3} (d = 5.24 g/cm^{3}). The density of Fe is 7.787 g/cm^{3}.

**Solution:**

1) Write the balanced chemical equation for the reaction:

2Fe +^{3}⁄_{2}O_{2}---> Fe_{2}O_{3}I decided to balance it with a fraction so as to use a Fe to Fe

_{2}O_{3}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 cm^{3} 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 Fe

_{2}O_{3}produced)

4) Convert moles of Fe_{2}O_{3} to grams:

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

5) Convert to cm^{3}:

11.1334 g / 5.24 g/cm^{3}= 2.12 cm^{3}The volume changes from 1.00 cm

^{3}to 2.12 cm^{3}.