Stoichiometry
Mass-Volume Problems #1 - 10

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Problem #1: This reaction was performed:

CaCO3(s) + 2HCl(aq) ---> CaCl2(s) + CO2(g) + H2O(ℓ)

What would be the volume of CO2 (at STP) produced from the complete reaction of 10.0 grams of CaCO2?

Solution:

1) Determine moles of CaCO3 reacted:

10.0 g / 100.086 g/mol = 0.099914 mol

2) Note that there is a 1:1 molar ratio between CaCO3 used and CO2 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:

(NH4)2Cr2O7(s) ---> N2(g) + 4H2O(g) + Cr2O3(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 N2 and H2O?

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 N2 and 4/5 is H2O.

partial pressure N2 ---> (0.030 atm) (0.2) = 0.0060 atm
partial pressure H2O ---> (0.030 atm) (0.8) = 0.024 atm

Problem #3: A 4.90-g sample of solid CoCl2 4H2O 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 CoCl2 4H2O in 4.90 g:

4.90 g / 201.8982 g/mol = 0.0242696 mol

2) Determine moles of water produced:

1 mol CoCl2 4H2O will produce 4 mol H2O

0.02427 mol CoCl2 4H2O will produce:

(4) (0.0242696 mol) = 0.0970784 mol H2O

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 H2 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 cmH2O = 112 mmH2O

(112 mmH2O) (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 H2.

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

n = 0.0015853 mol of H2 (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 ---> 2AlCl3 + 3H2

The key ratio is the Al to H2 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 #5: A 0.616 gram sample of a metal, M, reacts completely with sulfuric acid according to the reaction:

M(s) + H2SO4(aq) ---> MSO4(aq) + H2(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:

Ptotal = PH2 + PH2O

therefore:

1.0079 - 0.03167 = 0.97623 bar

2) Determine moles of H2 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 H2 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/mol

If the problem had asked to identify the metal, the answer would have been 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.

4NH3(g) + 5O2(g) ---> 4NO(g) + 6H2O(ℓ)

Solution:

1) Determine the limiting reagent:

ammonia: 0.46973 mol / 4 mol = 0.1174
oxygen: 0.34376 mol / 5 mol = 0.0687

Oxygen is the limiting reagent.

2) Determine moles of O2 present:

11.0 g / 31.9988 g/mol = 0.343763 mol

3) Determine moles of NO produced:

The molar ratio beween O2 and NO is 5 to 4.

5   0.343763 mol
––– = ––––––––––––
4   x

x = 0.2750104 mol of H2

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: C4H10 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:

2C4H10 + 13O2 ---> 8CO2 + 10H2O

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 O2 and H2O is 13 to 10.

13   x
––– = ––––––––––––
10   0.122688 mol

x = 0.1594944 mol of H2

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 H2 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 H2 that was collected?

Solution:

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

1) the H2 gas
2) water vapor
3) the 6.60 cm column of water

The 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/cm3) = (x) (13.534 g/cm3)

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 H2

4) Now, we use PV = nRT to determine the moles of H2:

(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 ---> ZnCl2 + H2

The molar ratio of Zn to H2 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 H2 gas at STP?

Solution to (b):

1) The chemical reaction is this:

2Na + 2HCl ---> 2NaCl + H2

In order to determine the amount of sodium that reacted, I must know how many moles of H2 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:

PH2 + 19.8 torr = 748.0 torr

PH2 = 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:

P1 = 728.2 torr      P2 = 760.0 torr     
V1 = 127.0 mL     V2 = x     
T1 = 295 K     T2 = 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 H2 to determine moles of Na that react:

2 is to 1 as x is to 0.00502675 mol

x = 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:

2NaN3 ---> 2Na + 3N2

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 cm3 = 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¯11) (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 NaN3

(0.733617 mol) (65.011 g/mol) = 47.7 g


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

2HClO3 + 4H2C2O4 ---> Cl2O + 8CO2 + 5H2O

Determine how many liters of carbon dioxide are obtained under normal operating conditions when 42.24 g HClO3 is reacted with 18.00 g H2C2O4.

Comment: this is a limiting reagent problem. I propose to solve the problem completely, twice. First, acting as if HClO3 is limiting and, second, acting as if the H2C2O4 is limiting.

Solution using HClO3:

1) Determine moles of HClO3:

42.24 g / 84.4579 g/mol = 0.500131 mol

2) Determine moles of CO2 produced:

1   0.500131 mol
––– = ––––––––––––
4   x

x = 2.000524 mol of CO2

Notice that I used a reduced ratio of 14 (the coefficients show 28 ).

3) Determine volume of CO2 produced at normal operating conditions:

PV = nRT

(1.00 atm) (V) = (2.000524 mol) (0.08206 L atm mol¯11) (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 H2C2O4:

1) Determine moles of H2C2O4:

18.00 g / 90.0338 g/mol = 0.1999249 mol

2) Determine moles of CO2 produced:

1   0.1999249 mol
––– = ––––––––––––
2   x

x = 0.3998498 mol of CO2

Notice that I used a reduced ratio of 12 (the coefficients show 48 ).

3) Determine volume of CO2 produced at normal operating conditions:

PV = nRT

(1.00 atm) (V) = (0.3998498 mol) (0.08206 L atm mol¯11) (298 K)

V = 9.78 L

The lesser volume produced shows the H2C2O4 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:

HClO3 ---> 0.500131 / 2 = 0.25
H2C2O4 ---> 0.1999249 / 4 = 0.05

The smaller value shows H2C2O4 to be the limiting reagent.


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