### StoichiometryMass-VolumeFifteen Examples

Example #1: What weight of CuO can be produced from 25.0 L. of O2 at STP by this reaction:

2Cu + O2 ---> 2CuO

Solution using molar volume:

1) Determine moles of O2:

25.0 L / 22.414 L/mol = 1.11537 mol

2) Determine moles of CuO produced:

The CuO to O2 molar ratio is 2 to 1.

2 is to 1 as x is to 1.11537 mol

x = 2.23074 mol

3) Determine mass of CuO produced:

(2.23074 mol) (79.545 g/mol) = 177.4 g (to three sig figs the answer is 177)

4) The above calculation using a dimensional analysis approach:

 25.0 L O2 1 mol O2 2 mol CuO 79.545 g CuO –––––––– x –––––––––– x ––––––––– x ––––––––––– = 177.4 g CuO 1 22.414 L O2 1 mol O2 1 mol CuO

Solution using the Ideal Gas Law:

1) Determine moles of O2:

PV = nRT

(1.00 atm) (25.0 L) = (n) (0.08206 L atm / mol K) (273 K)

n = 1.115953 mol

2) Determine moles of CuO produced:

The CuO to O2 molar ratio is 2 to 1.

2 is to 1 as x is to 1.115953 mol

x = 2.231906 mol

3) Determine mass of CuO produced:

(2.231906 mol) (79.545 g/mol) = 177.5 g (to three sig figs the answer is 178)

Example #2: Balance the equation and calculate the volume of O2 required to produce 75.0 g of P2O5 at STP.

P4 + O2 ---> P2O5

Solution:

1) Balanced equation:

P4 + 5O2 ---> 2P2O5

2) Determine moles of P2O5 to be produced:

75.0 g / 141.943 g/mol = 0.52838 mol

3) Determine moles of O2 required:

O2 and P2O5 are in a 5:2 molar ratio

 5 x –––– = –––––––––– 2 0.52838 mol

x = 1.32095 mol O2

4) Use PV = nRT to get volume of O2:

(1.00 atm) (x) = (1.32095 mol) (0.08206 L atm / mol K) (273 K)

x = 29.6 L

5) A dimensional analysis set up which uses molar volume:

 75.0 g P2O5 1 mol P2O5 5 mol O2 22.414 L O2 –––––––– x –––––––––––– x –––––––– x –––––––––– = 29.6 L O2 1 141.943 g P2O5 2 mol P2O5 1 mol O2

Example #3: 85.0 g of HI will produce how many liters of gaseous products at STP?

2HI ---> H2 + I2

Solution:

1) Determine moles of HI that decomposed:

85.0 g / 127.9079 g/mol = 0.66454 mol

2) There is a 1:1 molar ratio between HI and the moles of gaseous products (H2 and I2 added together). Therefore:

0.66454 mol of gaseous products

3) Use molar volume:

(0.66454) (22.414 L/mol) = 14.9 L

Example #4: What volume of H2 at STP is required to produce 93.0 g. of tungsten?

WO3 + 3H2 ---> W + 3H2O

Solution:

1) Determine moles of tungsten produced:

93.0 g / 183.84 g/mol = 0.505875 mol

2) Hydrogen gas and tungsten are in a 3:1 molar ratio:

 3 x –––– = –––––––––– 1 0.505875 mol

x = 1.517625 mol H2

3) Use PV = nRT to calculate the volume:

(1.00 atm) (V) = (1.517625 mol) (0.08206 L atm / mol K) (273 K)

V = 34.0 L

Example #5: What weight of dilute sulfuric acid (25% H2SO4 by weight) is required to produce 37.3 liters of hydrogen gas at STP?

Zn + H2SO4 ---> ZnSO4 + H2

Solution:

1) Determine moles of H2 at STP:

PV = nRT

n = PV / RT

n = [(1.00 atm) (37.3 L)] / [(0.08206 L atm/mol K) (273 K)]

n =1.665 mol

2) H2SO4 and H2 are in a 1:1 molar ratio based on the above balanced equation. Therefore:

1.665 mol of H2SO4 reacted

3) Determine grams of H2SO4 that reacted:

(1.665 mol ) (98.0768 g/mol) = 163.298 g

4) Determine mass of 25% solution required to deliver 163.298 g of H2SO4:

25% is to 100% as 163.298 g is to x

x = 653 g (to three sig figs)

Example #6: How many liters of O2 gas measured at 782.0 mmHg at 25.0 °C are required to completely react with 2.40 mol Al?

Solution:

1) Write the balanced chemical reaction:

4Al + 3O2 ---> 2Al2O3

2) Determine the moles of O2 required to react with 2.40 mol of Al:

the molar ratio for Al and O2 is 4 to 3, so we set up the following ratio and proportion:

 4 2.40 mol ––– = –––––––– 3 x

x = 1.80 moles of O2 required

3) Determine volume of O2 at stated P and T:

PV = nRT

V = nRT / P

x = [(1.80 mol) (0.08206 L atm / mol K) (298 K)] / 1.029 atm

x = 42.8 L

Comment: I converted 782.0 mmHg to atm.

Example #7: What volume of carbon dioxide, at 1.00 atm and 112.0 °C, will be produced when 80.0 grams of methane (CH4) is burned?

### Determine volume of CO2 produced

If we wished to determine the volume of O2 required, we would use 9.9732 mol. This is because of the 1:2 molar ratio between methane and oxygen. To burn 4.9866 mol of methane requires 9.9732 mol of oxygen. Using PV = nRT, this is 315 L (I doubled the 157.5 figure from the video, not the 158.)

Example #8: Propane, C3H8 reacts completely with oxygen to form carbon dioxide and water vapor. If 1.50 mole of propane is reacted with an excess of oxygen and the water vapor is collected and measured at 546 K and 1.00 atm, what volume of water vapor will be collected?

Solution:

1) Write a balanced chemical equation:

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

2) Determine moles of water vapor produced:

the molar ratio between propane used and water vapor produced is 1 to 4

therefore, water vapor will be produced in the following ratio and proportion:

1 is to 4 as 1.50 is to x

x = 6.00 moles of water vapor produced

3) Determine the volume of water vapor at the given temperature and pressure:

PV = nRT

V = nRT / P

x = [(6.00 mol) (0.08206 L atm mol¯11) (546 K)] / 1.00 atm

x = 269 L (to 3 sig. fig.)

Example #9: Oxygen gas is sometimes prepared in labs by the thermal decomposition of potassium chlorate (KClO3). The balanced chemical equation is as follows:

2KClO3(s) ---> 2KCl(s) + 3O2(g)

If 5.150 grams decompose, what volume of O2 would be obtained at STP?

Solution:

1) Determine moles of KClO3:

5.150 g / 122.6 g mol¯1 = 0.042006525 mol (I kept some guard digits.)

2) Determine moles of O2 produced:

the molar ratio of KClO3 used to O2 produced is 2 to 3

therefore, oxygen will be produced in the following ratio and proportion:

2 is to 3 as 0.042006525 is to x

x = 0.063009788 moles of O2 produced

3) Determine the volume of O2 produced at STP:

PV = nRT

V = nRT / P

x = [(0.063009788 mol) (0.08206 L atm mol¯11) (273 K)] / 1.000 atm

x = 1.412 L (to 4 sig. fig.)

Example #10: When heated to high temperatures, silver oxide (Ag2O) decomposes to form solid silver and oxygen gas. Calculate the volume of oxygen produced at STP by the decomposition of 7.44 g of Ag2O.

### Determine volume of O2 produced

Example #11: Given 21.45 L of oxygen gas at STP, how many grams of iron(III) oxide would be produced with excess iron?

Solution:

1) Write the chemical reaction:

2Fe + 32O2 ---> Fe2O3

The 1.5 to 1 molar ratio (I'll use 3 to 2 just below. Just to be different.) between O2 and Fe2O3 is the ratio to use.

2) Determine the moles of oxygen present. You can use PV = nRT or molar volume. I'll use molar volume:

 21.45 L ––––––––––– = 0.95699 mol 22.414 L/mol

The fact that the oxygen gas is at STP allows me to use molar volume. If the conditions had been not at STP, use of PV = nRT would have been required.

3) Using the O2 to Fe2O3 molar ratio referenced above, determine moles of Fe2O3 produced:

 3 0.95699 mol ––– = ––––––––––– 2 x

x = 0.6379933 mol [of iron(III) oxide produced]

4) Determine mass of iron(III) oxide produced:

(0.6379933 mol) (159.687 g/mol) = 101.9 g

Example #12: 1.10 g of sodium superoxide, NaO2, was reacted with excess water. Calculate the volume of oxygen gas measured at RTP produced in the reaction.

Solution:

1) Write the chemical reaction:

2NaO2 + H2O ---> 2NaOH + 32O2

I have deliberately left it balanced with a fractional coefficient.

2) Determine moles of NaO2 that reacted:

 1.10 g ––––––––– = 0.0200 mol 54.988 g/mol

3) NaO2 and O2 are in a 4:3 molar ratio (I left it as 2 to 1.5 up above and doubled it so as to use a whole-number ratio). Determine the moles of O2 produced:

4 is to 3 as 0.0200 mol is to x

x = 0.0150 mol

4) RTP stands for room temperature and pressure (see the next example for some discussion about RTP). The values I will use are these:

25.0 °C
1.00 atm

5) Determine the volume of 0.0150 mole of oxygen gas at RTP:

PV = nRT

(1.00 atm) (V) = (0.0150 mol) (0.08206 L atm / mol K) (298 K)

V = 0.3668 L

To three sig figs, this is 367 mL

Example #13: When 4.73 g of a solid was heated, the residue weighed 4.10 g and 320. cm3 of a gas (measured at room temperature) was evolved. Calculate the molecular mass of the gas.

Solution:

1) Determine the mass of gas produced:

4.73 g − 4.10 g = 0.63 g

Yay for the Law of Conservation of Mass!

2) Determine the moles of gas produced:

PV = nRT

(____) (0.320 L) = (n) (0.08206 L atm / mol K) (298 K)

A little bit of discussion: (a) the problem is completely silent about the pressure, but we can take our cue from 'room temperature' and use a value of 1.00 atm for 'room pressure' and (b) note my use of 298 K means that 'room temperature' is 25.0 °C.

Room temperature and pressure is often abbreviated RTP and it NOT a standard thing in chemistry. Some people use 20.0 °C for RT and some use 1 bar for RP. Make sure you consult with your teacher as to the RTP values they wish to be used.

(1.00 atm) (0.320 L) = (n) (0.08206 L atm / mol K) (298 K)

n = 0.013086 mol

3) Determine the molar mass of the gas:

0.63 g / 0.013086 mol = 48.14 g/mol

4) Sometimes an instructor will lump steps 2 and 3 together. Like this:

Use PV = nRT and this

n = mass / MW (where MW is the molecular weight or molar mass of the substance)

We can substitute the second equation into PV = nRT thus:

PV = (mass / MW) RT

rearrange:

MW = (mass / V) x (RT/P)

using the values from the problem:

 0.63 g (0.08206 L atm/mol K) (298 K) MW = ––––––– x ––––––––––––––––––––––––– = 48.14 g/mol 0.320 L 1.00 atm

Example #14 What volume of hydrogen gas will be produced when 34.7 g of zinc react with excess hydrochloric acid at SATP?

Solution:

1) The chemical equation:

Zn + 2HCl ---> ZnCl2 + H2

The key molar ratio is between Zn and H2 and it is a 1:1 ratio.

2) Determine moles of Zn:

34.7 g / 65.38 g/mol = 0.530743 mol

3) Based on the 1:1 molar ratio above, we determine that:

0.530743 mol of H2 is produced.

4) The conditions of SATP (Standard Ambient Temperature and Pressure) are 25.0 °C and 100.0 kPa. What we must do now is determine the volume of H2 gas at SATP:

PV = nRT

(100.0 kPa / 101.325 kPa/atm) (V) = (0.530743 mol) (0.08206 L atm mol¯11 K) (298.0 K)

V = 13.2 L (to three sig figs)

The (100.0 kPa / 101.325 kPa/atm) factor converts kPa to atm, so I can use the R value I did.

(34.7 g Zn) / (65.38 g Zn/mol) x (1 mol H2 / 1 mol Zn) x (24.8 L/mol) = 13.2 L H2

Some discussion on how to determine the 24.8 L/mol:

The conditions for 22.414 L/mol are 0 °C (273.15 K) and 1.00 atm (101.325 kPa). We need to convert to 25.0 °C (298.15 K) and 100.0 kPa. We do that with the combined gas law:

 P1V1 P2V2 ––––– = ––––– T1 T2

and fill in the values:

 (101.325 kPa) (22.414 L) (100.0 kPa) (x) –––––––––––––––––––– = –––––––––––– 273.15 K 298.15 K

cross-multiply and divide for the answer:

x = 24.7896 L

to three sig figs, this is 24.8 L for the molar volume at STAP

Example #15: Solid lithium hydroxide is used to "scrub" CO2 from the air in spacecraft and submarines; it reacts with the CO2 to produce lithium carbonate and water. What volume in liters of CO2 at 23.0 °C and 722.0 torr can be removed by reaction with 478.0 g of lithium hydroxide?

Solution:

1) First, let us write the balanced chemical equation:

2LiOH + CO2 ---> Li2CO3 + H2O

The molar ratio of importance is the 2:1 ratio between LiOH and CO2

2) Convert 478.0 g to moles:

478.0 g / 23.9479 g/mol = 19.96 mol

3) Use the molar ratio above to determine moles of CO2 that react:

2 is to 1 as 19.96 ml is to x

x = 9.98 mol

4) Determine volume in liters:

PV = nRT

(722.0 torr / 760.0 torr/atm) (V) = (9.98 mol) (0.08206 L atm / mol K) (300. K)

V = 259 L (to three sig figs)

Bonus Example #1:

Bonus Problem: 12CO2(g) + 11H2O(ℓ) ---> C12H22O11(s) + 12O2(g)

(a) How many grams of sucrose (C12H22O11) are produced from 224 liters of carbon dioxide (CO2) at STP?
(b) How many liters of carbon dioxide (CO2) at STP are needed to produce 5.00 pounds of sucrose? (1 kg = 2.20462 lbs.)
(c) What mass of water would be needed to combine with 100. liters of CO2, measured at RTP?

Solution to (a):

1) Determine moles of CO2:

224 L / 22.414 L/mol = 9.993754 mol

2) Determine moles, then mass, of sucrose:

There is a 12:1 molar ratio between carbon dioxide and sucrose.

9.993754 mol / 12 = 0.832813 moles of sucrose

(0.832813 mol) (342.2948 g/mol) = 285 g

Solution to (b):

1) Convert pounds of sucrose to moles:

(5.00 lbs) (1 kg / 2.20462 lbs) (1000 g / 1 kg) (1 mol / 342.2948 g/mol) = 6.625764 mol

2) Determine moles of CO2 required:

There is a 12:1 molar ratio between CO2 and sucrose.

(6.625764 mol) (12) = 79.509168 mol

3) Determine volume of CO2:

PV = nRT

(1.00 atm) (V) = (79.509168 mol) (0.08206 L atm / mol K) (273 K)

V = 1781.1946 L (to three sig figs, rounded from 1781 L)

or

(22.414 L/mol) (79.509168 mol) = 1782.1185 L

to three sig figs, 1780 L

Solution to (c):

1) RTP means "room temperature and pressure." It is not a standard chemical term and there are two different sets of values used:

(a) 20.0 °C and 1.00 atm
(b) 25.0° and 1.00 atm

I will use (a), the one used by the National Institute of Standards and Technology (NIST).

2) Determine moles of CO2:

PV = nRT

(1.00 atm) (100. L) = (n) (0.08206 L atm / mol K) (293 K)

n = 4.15911 mol

3) Determine moles, then mass, of H2O:

There is a 12:11 molar ratio between CO2 and H2O.

12 is to 11 as 4.15911 is to x

x = 3.8125175 mol

(3.8125175 mol) (18.015 g/mol) = 68.7 g (to three sig figs)

Bonus Example #2: If 2.50 x 1024 atoms of iron are produced, how many liters of carbon monoxide are produced? The reaction is:

Fe2O3 + 3C ---> 2Fe + 3CO

Solution:

1) Convert atoms of iron to moles of iron:

2.50 x 1024 atoms / 6.022 x 1023 atoms/mole = 4.151445 mol

2) Use the molar ratio between Fe and CO:

 2 4.151445 mol ––– = ––––––––––– 3 x

x = 6.2271675 mol of CO produced

3) No pressure/temperature data provided in the problem. Assume conditions are STP:

(22.414 L/mol) (6.2271675 mol) = 140. L of CO at STP

4) You can also use PV = nRT to solve the problem:

(1.00 atm) (V) = (6.2271675 mol) (0.08206 L atm / mol K) (273 K)