Stoichiometry
Volume-Volume Examples
(volume of gas, not solution)

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Problem #1: If 3.62 dm3 of hydrogen gas is reacted, what volume of oxygen gas would be used at STP?

Solution:

1) The balanced chemical equation is:

2H2(g) + O2(g) ---> 2H2O(g)

2) A side note:

Notice how I do not write this equation:
H2(g) + O2(g) ---> H2O2(ℓ)

When hydrogen and oxygen react directly, water is ALWAYS formed. To make hydrogen peroxide, a different reaction (actually several) must be used.

3) All gases are at the same temperature and pressure (STP in this case), therefore the volumes of gases used are directly related to the molar ratios from the balanced equation. (Reminder: Avogadro's Hypothesis.) So:

2 mol H2   3.62 dm3
–––––––  =  –––––––
1 mol O2   x

x = 1.81 dm3

4) A repeat: If all gases are at the same temperature and pressure, the volumes of gases used are directly related to the molar ratios from the balanced equation. So:

(3.62 dm3 H2) (1 dm3 O2 / 2 dm3 H2) = 1.81 dm3 O2

Problem #2: Ammonia is often formed by reacting nitrogen and hydrogen gases. How many liters of ammonia gas can be formed from 24.5 L of hydrogen gas at 93.0 °C and a pressure of 33.9 kPa?

Solution:

1) Write the chemical equation:

N2(g) + 3H2(g) ---> 2NH3(g)

2) If the volume of ammonia produced is measured under the same pressure and temperature conditions as the hydrogen, then this relationship exists:

3 L of H2 produce 2 L NH3

3) Therefore:

3   24.5 L
–––  =  –––––
2   x

x = 16.3 L of NH3 produced


Problem #3: At STP, how many liters of NH3 can be produced form 28.0 L of N2 and 45.0 L of H2? Both reactant gases are at STP.

Solution:

1) The balanced chemical equation for this reaction is as follows:

N2 + 3H2 ---> 2NH3

2) Since everything is measured at a constant temperature and pressure, the volumes are in direct proportion to the number of moles involved. Since there are differing amounts of the two reactants, we need to determine the limiting reagent:

N2 ---> 28 / 1 = 28
H2 ---> 45 / 3 = 15

H2 is the limiting reagent. This means that the H2 to NH3 molar ratio of 3:2 will be used to determine the volume of NH3 produced.

3) Determine volume of NH3 produced:

3   45.0 L
–––––––  =  –––––––
2   x

x = 30.0 L (this is the answer to the question)

4) Instead of using the division technique in step 2 to determine the limiting reagent, you can do the calculation above with the other reactant. Here follows the calculation for N2:

1   28.0 L
–––––––  =  –––––––
2   x

x = 56.0 L

However, there is insufficient hydrogen to produce 56.0 L (its amount can only produce 30.0 L). From this, we conclude that nitrogen is not the limiting reagent, hydrogen is and that 30.0 L is the answer to the question.


Problem #4: How many liters of NH3 (at STP) can be produced form 28.0 L of N2 (at 750. mmHg and 30.0 °C) and 45.0 L of H2 (at 2.10 atm and 25.0 °C)?

Solution:

1) The balanced chemical equation for this reaction is as follows:

N2 + 3H2 ---> 2NH3

2) When the gases are at different pressures and temperatures, we have two directions we can follow:

(a) convert all volumes to the same pressure and temperature. The volumes would then be in direct proportion to the moles. This is the usual way when a volume is requested for the final answer.
(b) convert all volumes to moles (using PV = nRT, calculate moles of the gases asked for in the problem, then use PV = nRT to get the volume. This is the usual path if moles or mass of the unknown gas is requested.

I shall follow (a) in this problem.

3) Using the combined gas law, determine the volumes of N2 and H2 at STP. Here's the combined gas law equation:

P1V1   P1V2
––––– = –––––
T1   T2

4) Here's a combined gas law solution matrix filled in for nitrogen:

P1 = 750.0 mmHgP2 = 760.0 mmHg
V1 = 28.0 LV2 = _____
T1 = 303 KT2 = 273 K

V2 for nitrogen = 24.896 L

5) Here's a combined gas law solution matrix filled in for hydrogen:

P1 = 2.10 atmP2 = 1.00 atm
V1 = 45.0 LV2 = _____
T1 = 298 KT2 = 273 K

V2 for hydrogen = 103.154 L

6) Solve for the volume of ammonia produced as if nitrogen were the limiting reagent:

1   24.896 L
–––––––  =  –––––––
2   x

x = 49.792 L

7) Solve for the volume of ammonia produced as if hydrogen were the limiting reagent:

3   103.154 L
–––––––  =  –––––––
2   x

x = 68.769 L

8) Nitrogen is the limiting reagent and the answer to the problem is 49.8 L (to three sig figs).


Problem #5: What volume of O2 at 298 K and 2.00 atm is required to react completely with 2.00 L of methane at the same temperature and pressure? What volume of CO2 would be produced if it was measured at 10.0 atm and 300. K?

Solution:

1) A balanced chemical equation is always nice to have:

CH4 + 2O2 ---> CO2 + 2H2O

2) Because of the equal temperature and equal pressure stipulation, this means that the volumes react in the same ratio as the ratio given by the coefficients of the balanced equation. (Recall Avogadro's Hypothesis.)

1 is to 2 as 2.00 L is to x

x = 4.00 L of O2 required

3) Methane and carbon dioxide react in a 1:1 volume ratio:

1 is to 1 as 2.00 L is to x

x = 2.00 L of CO2 produced at 298 K and 2.00 atm

4) Convert CO2 volume just calculated to the conditions given in the problem. Use the Combined Gas Law:

P1V1   P2V2
––––– = –––––
T1   T2

(2.00 atm) (2.00 L)   (10.0 atm) (x)
––––––––––––––– = ––––––––––––
298 K   300. K

(298 K) (10.0 atm) (x) = (2.00 atm) (2.00 L) (300. K)

x = 0.403 L (to three sig figs)

5) Note the use of the Combined Gas Law. Typically, volumes in stoichiometry problms involve some use of the Ideal Gas Law. The above is an exception.


Problem #6: What volume of hydrogen gas at STP (101.325 kPa, 273 K) is required to react with 42.0 litres of oxygen gas at SATP (100.0 kPa, 298.0 K)?

Solution:

1) Write the chemical equation:

2H2 + O2 ---> 2H2O

The 2:1 molar ratio between H2 and O2 is the ratio we will use.

By the way, water is the only product when H2 and O2 react directly. H2O2 must be produced by other means. That's how I knew to write H2O for the product.

2) Let us determine the moles of O2 present:

PV = nRT

(100.0 kPa / 101.325 kPa/atm) (42.0 L) = (n) (0.08206 L atm mol¯11 K) (298.0 K)

n = 1.69506 mol

3) Because of the 2:1 molar ratio, 3.39012 mol of H2 is required. Determine its volume at STP:

PV = nRT

(1.00 atm) (V) = (3.39012 mol) (0.08206 L atm mol¯11 K) (273 K) V = 75.9 L

4) At this link are two other answers:

https://answers.yahoo.com/question/index?qid=20180125190131AALNNQ5

Both get at the answer in a different fashion than my solution. Note that dr W mentions the ideal Gas Law, but he uses a form of the Combined Gas Law (which, I admit, can be "derived" from the Ideal Gas Law).


Problem #7: Gas barbecues burn propane using oxygen from the air. If 3.50 L of propane is burned while you are cooking burgers (yum!), what volume of oxygen is required for combustion of the propane at a constant temperature and pressure?

Solution:

1) Write the chemical equation for propane combusting:

C3H8 + 5O2 ---> 3CO2 + 4H2O

2) Since everything occurs at constant T and P, all consideration of T and P drops away. Remember, we would only have to consider T or P if either changed. See below.

3) Since P and T remain constant in this problem, we can do this:

PV = nRT

V = (RT/P) (n)

4) Since (RT/P) is a constant, the volumes of gas are in direct proportion to the number of moles. To solve the problem:

1 is to 5 as 3.50 is to x

x = 17.5 L

By the way, that's 17.5 L of pure oxygen gas. If, perchance, you were asked to determine the total volume of atmospheric air needed to supply 17.5 L of pure oxygen gas at STP, the calculation would be 17.5 / 0.20 = 87.5 L. This is because oxygen gas is approximately 20% of the atmosphere.

Regarding T and P both changing, consider this problem:

3.50 L of propane at 25.0 °C and 745.0 mmHg pressure would require how many liters of O2 (measured at STP) for complete combustion.

In that problem, we would calculate moles of C3H8, then determine moles of O2 and then convert moles to volume at STP. Two usages of PV = nRT would be required. As follows:

(745.0 mmHg / 760.0 mmHg/atm) (3.50 L) = (n) (0.08206 L atm / mol K) (298.0 K)

n = 0.1403 mol of C3H8

0.7015 mol of O2 is required (1 mole of C3H8 requires 5 moles of O2)

(1.00 atm) (V) = (0.7015 mol) (0.08206 L atm / mol K) (273.0 K)

V = 15.7 L

Note: you could also use molar volume (since the final conditions are at STP):

(22.414 L/mol) (0.7015 mol = 15.7 L


Problem #8: A mixture of hydrogen and oxygen (45 mL) is sparked to form liquid water. The component not in excess reacts completely and 15 mL is left over. (All measurements are made at the same temperature and pressure). The composition by volume in the original mixture of H2 : O2 is:

Solution:

1) Let us assume the excess is H2.

15 mL of H2 left over

Of the other 30 mL:

20 mL = H2 and 10 mL = O2

Original mix was 35 mL H2 and 10 mL O2, that's a 7:2 ratio

2) Let us assume the excess is O2.

15 mL of O2 left over

Of the other 30 mL:

20 mL = H2 and 10 mL = O2

Original mix was 20 mL H2 and 25 mL O2, that's a 4:5 ratio


Problem #9: What volume of oxygen gas is needed to react completely with 0.587 L of carbon monoxide gas (CO) to form gaseous carbon dioxide?

Solution:

1) Write the balanced chemical reaction:

2CO + O2 ---> 2CO2

Note the 2:1 molar ratio between CO and O2.

2) No information is given in the problem concerning pressure and temperature. Therefore, assume that the amounts of CO and O2 are measured at the same pressure and the temperature.

3) This allows Avogadro's Hypothesis (equal volumes of gases at equal P and T contain equal number of gas molecules) to be invoked. The consequence of this is that the 2:1 molar ratio is also (at equal P and T mind you!) is also a 2:1 volumetric ratio.

4) Solve:

2   0.578 L
–––  =  ––––––––––
1   x

x = 0.289 L


Problem #10: When 0.020 mole of a hydrocarbon X was completely burnt in oxygen, 960. mL of carbon dioxide was formed at room temperature and pressure. (a) Calculate the volume of carbon dioxide formed at RTP from one mole of hydrocarbon X. (b) Calculate the number of carbon atoms in one molecule of hydrocarbon X.

Solution:

1) Calculate volume of CO2 (aT RTP) given off by one mole of X combusting:

960. mL / 0.020 mol = 48000 mL = 48.0 L <--- answer to (a)

2) Take RTP to be equal to 1.00 atm and 20.0 °C

Sometimes, RTP is taken to be 1.00 atm and 25.0 °C. This is because RTP is not a standardized term in chemistry.

3) Calculate molar volume at RTP:

PV = NRT

(1.00 atm) (V) = (1.00 mol) (0.08206 L atm / mol K) (293 K)

V = 24.0 L

4) Determine moles of C in one mole of X:

48 L / 24 L/mol = 2

Therefore, 2 atoms of C per one molecule of X <--- answer to (b)


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