Mass-Volume Problems

#1 - 10

**Problem #1:** This reaction was carried out:

CaCO_{3}(s) + 2HCl(aq) ---> CaCl_{2}(s) + CO_{2}(g) + H_{2}O(ℓ)

What would be the volume of CO_{2} (at STP) produced from the complete reaction of 10.0 grams of CaCO_{2}?

**Solution:**

1) Determine moles of CaCO_{3} reacted:

10.0 g / 100.086 g/mol = 0.099914 mol

2) Note that there is a 1:1 molar ratio between CaCO_{3} used and CO_{2} produced.

3) Since the problem is using STP, we can use molar volume:

(22.414 L/mol) (0.099914 mol) = 2.24 L (to three sig figs)

4) If the pressure and temperature were not at STP, we would use the ideal gas law to calculate the volume produced.

PV = nRT(1.00 atm) (V) = (0.099914 mol) (0.08206 L atm / mol K) (273 K)

V = 2.24 L (to three sig figs)

Note that I set up the problem using STP values. If non-STP values were to be used, they are typically given in the body of the question.

**Problem #2:** 0.84 g of ammonium dichromate is decomposed. Here is the chemical reaction:

(NH_{4})_{2}Cr_{2}O_{7}(s) ---> N_{2}(g) + 4H_{2}O(g) + Cr_{2}O_{3}(s)

The gases from this reaction are trapped in a 13.6 L flask at 26.0 °C. (a) What is the total pressure of the gas in the flask? (b) What are the partial pressures of N_{2} and H_{2}O?

**Solution:**

1) Determine the moles of ammonium dichromate used:

0.84 g / 252.0622 g/mol = 0.00333251 mol

2) From the balanced equation, each mole of ammonium dichromate decomposed produces 5 moles of gas.

0.00333251 mol x (5 mol gas / 1 mol dichromate) = 0.01666255 mol gas

3) Use the ideal gas law to calculate the pressure:

PV = nRT(P) (13.6 L) = (0.01666255 mol) (0.08206 L atm/mol K) (299.0 K)

P = 0.030 atm (to two sig figs, based on the 0.84)

4) Of the gases produced, 1/5 is N_{2} and 4/5 is H_{2}O.

partial pressure N_{2}---> (0.030 atm) (0.2) = 0.0060 atm

partial pressure H_{2}O ---> (0.030 atm) (0.8) = 0.024 atm

**Problem #3:** A 4.90-g sample of solid CoCl_{2} **⋅** 4H_{2}O was heated such that the water turned to steam and was driven off. Assuming ideal behavior, what volume would that steam occupy at 1.00 atm and 100.0 °C?

**Solution:**

1) Determine moles of CoCl_{2} **⋅** 4H_{2}O in 4.90 g:

4.90 g / 201.8982 g/mol = 0.0242696 mol

2) Determine moles of water produced:

1 mol CoCl_{2}⋅4H_{2}O will produce 4 mol H_{2}O0.02427 mol CoCl

_{2}⋅4H_{2}O will produce:(4) (0.0242696 mol) = 0.0970784 mol H

_{2}O

3) Calculate volume at 1.00 atm and 100.0 °C:

PV = nRT(1.00 atm) (V) = (0.0970784 mol) (0.082057 L atm / mol K) (373 K)

V = 2.97 L (to three sig figs)

**Problem #4:** If 39.5 mL of H_{2} are produced at 21.0 °C when the atmospheric pressure is 762.8 mmHg, and the height of the liquid column in the eudiometer is 11.2 cm, what mass of aluminum is used?

**Solution:**

1) The pressure of the wet gas in the eudiometer plus the 11.2 cm of water equals the measured atmospheric pressure. We need to obtain the pressure of the dry gas.

11.2 cmH_{2}O = 112 mmH_{2}O(112 mmH

_{2}O) (1.00 g/mL) = (mmHg) (13.6 g/mL)x = 8.2353 mmHg

At 21.0 °C, the vapor pressure of water is 18.7 mmHg. Found here.

762.8 − 8.2353 − 18.7 = 735.8647 mmHg

2) The next step is to use PV = nRT to determine moles of H_{2}.

(735.8647 mmHg / 760 mmHg/atm) (0.0395 L) = (n) (0.08206 L atm / mol K) (294 K)n = 0.0015853 mol of H

_{2}(notice that I have carried several guard digits)

3) Let's write a chemical equation. Let us assume the Al reacted with HCl.

2Al + 6HCl ---> 2AlCl_{3}+ 3H_{2}The key ratio is the Al to H

_{2}molar ratio of 2 to 3

2 x ––– = –––––––––––– 3 0.0015853 mol y = 0.00105687 mol of Al

4) The last step:

(0.00105687 mol) (26.98 g/mol) = 0.0285 g (to three sig figs)

**Problem #5a:** A 0.616 gram sample of a metal, M, reacts completely with sulfuric acid according to the reaction:

M(s) + H_{2}SO_{4}(aq) ---> MSO_{4}(aq) + H_{2}(g)

A volume of 239 mL of hydrogen is collected over water; the water level in the collecting vessel is the same as the outside level. Atmospheric pressure is 1.0079 bar and the temperature is 25.0 °C. The vapor pressure of water at 25.0 °C is 0.03167 bar. Calculate the molar mass of the metal.

**Solution:**

1) Use Dalton's Law to get the pressure of the dry hydrogen:

P_{total}= P_{H2}+ P_{H2O}therefore:

1.0079 − 0.03167 = 0.97623 bar

2) Determine moles of H_{2} produced:

PV = nRT(0.97623 bar) (0.239 L) = (n) (0.0831447 L bar / mol K) (298 K)

n = 0.00941671 mol

Note that R has the units L bar / mol K. This is a lesser-used unit for R, but the value associated with that unit can be easily looked up.

3) From the balanced chemical equation, M and H_{2} are in a 1:1 molar ratio. Therefore:

0.00941671 mole of M was consumed.

4) Calculate the molar mass of M:

0.616 g / 0.00941671 mol = 65.4 g/molIf the problem had asked to identify the metal, the answer would have been zinc.

**Problem #5b:** 19.4 grams of metal sulfide has reacted with H_{2}SO_{4} solution and 4.46 L of H_{2}S gas was collected at STP. The oxidation number of the metal is +2. Determine what element the metal is.

**Solution:**

MS + H_{2}SO_{4}---> H_{2}S + M_{2}SO_{4}[balanced as written](4.46 L) / (22.414 L/mol) (1 mol MS / 1 mol H

_{2}S) = 0.1989828 mol MS(19.4 g MS) / (0.1989828 mol MS) = 97.495864 g /mol <--- the molar mass of MS

(97.495864 g/mol) − (32.0655 g/mol) = 65.43 g/mol <--- 32.0655 g/mol is the atomic weight of sulfur

65.43 g/mol is (within experimental error) the atomic weight of zinc.

**Problem #6:** Calculate the volume of nitrogen monoxide gas produced when 8.00 g of ammonia is reacted with 11.0 g of oxygen at 25.0 °C. The density of nitrogen monoxide at 25.0 °C is 1.23 g/L.

4NH_{3}(g) + 5O_{2}(g) ---> 4NO(g) + 6H_{2}O(ℓ)

**Solution:**

1) Determine the limiting reagent:

ammonia: 0.46973 mol / 4 mol = 0.1174

oxygen: 0.34376 mol / 5 mol = 0.0687Oxygen is the limiting reagent.

2) Determine moles of O_{2} present:

11.0 g / 31.9988 g/mol = 0.343763 mol

3) Determine moles of NO produced:

The molar ratio beween O_{2}and NO is 5 to 4.

5 0.343763 mol ––– = –––––––––––– 4 x x = 0.2750104 mol of H

_{2}

4) Determine mass of NO produced:

0.2750104 mol times 30.006 g/mol = 8.251962 g

5) Determine volume of NO produced:

8.251962 g divided by 1.23 g/L = 6.61 L

**Problem #7:** C_{4}H_{10} combusts. What mass of oxygen is needed to make 3.00 L of water at 0.990 atm and 295 K.

**Solution:**

1) The balanced equation:

2C_{4}H_{10}+ 13O_{2}---> 8CO_{2}+ 10H_{2}O

2) Moles of water present in the 3.00 L:

PV = nRT(0.990 atm) (3.00 L) = (n) (0.08206 L atm / mol K) (295 K)

n = 0.122688 mol

3) The molar ratio between O_{2} and H_{2}O is 13 to 10.

13 x ––– = –––––––––––– 10 0.122688 mol x = 0.1594944 mol of H

_{2}

4) Moles to grams:

(0.1594944 mol) (32.0 g/mol) = 5.10 g (to three sig figs)

**Problem #8:** A student collected 17.32 mL of H_{2} over water at 30.0 °C. The water level inside the collection apparatus was 6.60 cm higher than the water level outside. The barometric pressure was 731.0 torr. How many grams of zinc had to react with the HCl solution to produce the H_{2} that was collected?

**Solution:**

1) The pressure inside the collection tube is made up of three things:

1) the H_{2}gas

2) water vapor

3) the 6.60 cm column of waterThe sum of the three above pressures is equal to 731 torr.

2) The 6.60 cm column of water must be converted to mmHg pressure.

We use the densities of water and mercury to do this:(6.60 cm) (1.00 g/cm

^{3}) = (x) (13.534 g/cm^{3})x = 0.488 cm of mercury

This equals 4.88 mmHg

731.0 − 4.88 = 726.12 mmHg <--- the pressure of the gas minus the 6.60 cm column of water

3) Now, we need to remove the pressure of the water vapor.

We need the vapor pressure of water at 30.0 °C, a value we look up.726.12 − 31.8 = 694.32 mmHg <--- that's the pressure of just the H

_{2}

4) Now, we use PV = nRT to determine the moles of H_{2}:

(694.32 / 760) (0.01732 L) = (n) (0.08206 L atm / mol K) (303 K)n = 0.0006364 mol

(694.32 / 760) <--- that converts mmHg to atm

5) Now, some stoichiometry to get the mass of zinc:

Zn + 2HCl ---> ZnCl_{2}+ H_{2}The molar ratio of Zn to H

_{2}is 1:1, so we now know that 0.0006364 mol of Zn was used.0.0006364 mol times 65.38 g/mol = 0.0416 g

**Problem #9:** Hydrogen gas is produced by the reaction of sodium metal with an excess of hydrochloric acid solution. The hydrogen gas was collected by water displacement at 22.0 °C and 127.0 mL was collected with a total pressure of 748.0 torr. The vapor pressure of water at 22.0 °C is 19.8 torr.

(a) What mass of sodium metal was consumed in the reaction?

(b) What is the volume of dry H_{2}gas at STP?

**Solution to (b):**

1) The chemical reaction is this:

2Na + 2HCl ---> 2NaCl + H_{2}

In order to determine the amount of sodium that reacted, I must know how many moles of H_{2} was produced. I do that by solving (b) first.

2) Use Dalton's Law of Partial Pressures to determine the pressure of the dry hydrogen gas:

P_{H2}+ 19.8 torr = 748.0 torrP

_{H2}= 728.2 torr

3) Use PV = nRT to determine moles of gas:

(728.2 torr / 760.0 torr/atm) (0.1270 L) = (n) (0.08206 L atm / mol K) (295 K)n = 0.00502675 mol

I used the Ideal Gas Law because I knew I needed moles of hydrogen to get my answer for (a).

4) Continue with the solution for (b) by using the Combined Gas Law to get the volume of gas at STP:

P _{1}= 728.2 torrP _{2}= 760.0 torrV _{1}= 127.0 mLV _{2}= xT _{1}= 295 KT _{2}= 273 K

(728.2 torr) (127.0 mL) (760.0 torr) (x) –––––––––––––––––– = ––––––––––– 295 K 273 K x = 112.6 mL

**Solution to (a):**

1) Use 2:1 molar ratio between Na and H_{2} to determine moles of Na that react:

2 is to 1 as x is to 0.00502675 molx = 0.0100535 mol

2) Convert moles to grams:

(0.0100535 mol) (22.99 g/mol) = 0.2311 g

**Problem #10:** Automobile air bags inflate during a crash or sudden stop by the rapid generation of nitrogen gas from sodium azide according to the reaction:

2NaN_{3}---> 2Na + 3N_{2}

How many grams of sodium azide are needed to provide sufficient nitrogen gas to fill a 30.0 cm x 30.0 cm x 25.0 cm bag to a pressure of 1.20 atm at 26.0 °C?

**Solution:**

1) Determine the volume of the bag:

30.0 cm x 30.0 cm x 25.0 cm = 22500 cm^{3}= 22500 mL = 22.5 L

2) Determine moles of gas filling the bag:

PV = nRT(1.20 atm) (22.5 L) = (n) (0.08206 L atm mol¯

^{1}K¯^{1}) (299 K)n = 1.100426 mol

3) Determine moles, then mass of sodium azide required:

2 x ––– = –––––––––––– 3 1.100426 mol x = 0.733617 mol of NaN

_{3}(0.733617 mol) (65.011 g/mol) = 47.7 g

**Bonus Problem:** Chloric acid reacts with oxalic acid. Here is one possible reaction:

2HClO_{3}+ 4H_{2}C_{2}O_{4}---> Cl_{2}O + 8CO_{2}+ 5H_{2}O

Determine how many liters of carbon dioxide are obtained under normal operating conditions when 42.24 g HClO_{3} is reacted with 18.00 g H_{2}C_{2}O_{4}.

Comment: this is a limiting reagent problem. I propose to solve the problem completely, twice. First, acting as if HClO_{3} is limiting and, second, acting as if the H_{2}C_{2}O_{4} is limiting.

**Solution using HClO _{3}:**

1) Determine moles of HClO_{3}:

42.24 g / 84.4579 g/mol = 0.500131 mol

2) Determine moles of CO_{2} produced:

1 0.500131 mol ––– = –––––––––––– 4 x x = 2.000524 mol of CO

_{2}Notice that I used a reduced ratio of

^{1}⁄_{4}(the coefficients show^{2}⁄_{8}).

3) Determine volume of CO_{2} produced at normal operating conditions:

PV = nRT(1.00 atm) (V) = (2.000524 mol) (0.08206 L atm mol¯

^{1}K¯^{1}) (298 K)V = 48.9 L

Notice that I took "normal operating conditions" to be RTP (room temperature and pressure). Just remember, RTP is not a standard chemical thing, some people use 20.0 °C for the room temperature.

**Solution using H _{2}C_{2}O_{4}:**

1) Determine moles of H_{2}C_{2}O_{4}:

18.00 g / 90.0338 g/mol = 0.1999249 mol

2) Determine moles of CO_{2} produced:

1 0.1999249 mol ––– = –––––––––––– 2 x x = 0.3998498 mol of CO

_{2}Notice that I used a reduced ratio of

^{1}⁄_{2}(the coefficients show^{4}⁄_{8}).

3) Determine volume of CO_{2} produced at normal operating conditions:

PV = nRT(1.00 atm) (V) = (0.3998498 mol) (0.08206 L atm mol¯

^{1}K¯^{1}) (298 K)V = 9.78 L

The lesser volume produced shows the H

_{2}C_{2}O_{4}to be the limiting reagent and, thus, 9.78 L is the answer to the question.

Comment: the limiting reagent could have been determined thusly:

HClO_{3}---> 0.500131 / 2 = 0.25

H_{2}C_{2}O_{4}---> 0.1999249 / 4 = 0.05The smaller value shows H

_{2}C_{2}O_{4}to be the limiting reagent.