Mass-Volume Examples

**Example #1:** How many liters of O_{2} 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 + 3O_{2}---> 2Al_{2}O_{3}

2) Determine the moles of O_{2} required to react with 2.40 mol of Al:

the molar ratio for Al and O_{2}is 4 to 3, so we set up the following ratio and proportion:

4 2.40 mol ––– = –––––––– 3 x x = 1.80 moles of O

_{2}required

3) Determine volume of O_{2} at stated P and T:

PV = nRTV = 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 #2:** What volume of carbon dioxide, at 1.00 atm and 112.0 °C, will be produced when 80.0 grams of methane (CH_{4}) is burned?

Video: the solution to the above problem

If we wished to determine the volume of O

_{2}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 #3:** Propane, C_{3}H_{8} 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:

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

2) Determine moles of water vapor produced:

the molar ratio between propane used and water vapor produced is 1 to 4therefore, water vapor will be produced in the following ratio and proportion:

1 is to 4 as 1.50 is to xx = 6.00 moles of water vapor produced

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

PV = nRTV = nRT / P

x = [ (6.00 mol) (0.08206 L atm mol

^{-1}K^{-1}) (546 K) ] / 1.00 atmx = 269 L (to 3 sig. fig.)

**Example #4:** Oxygen gas is sometimes prepared in labs by the thermal decomposition of potassium chlorate (KClO_{3}). The balanced chemical equation is as follows:

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

If 5.150 grams decompose, what volume of O_{2} would be obtained at STP?

**Solution:**

1) Determine moles of KClO_{3}:

5.150 g / 122.6 g mol^{-1}= 0.042006525 mol (I kept some guard digits.)

2) Determine moles of O_{2} produced:

the molar ratio of KClO_{3}used to O_{2}produced is 2 to 3therefore, oxygen will be produced in the following ratio and proportion:

2 is to 3 as 0.042006525 is to xx = 0.063009788 moles of O

_{2}produced

3) Determine the volume of O_{2} produced at STP:

PV = nRTV = nRT / P

x = [ (0.063009788 mol) (0.08206 L atm mol

^{-1}K^{-1}) (273 K) ] / 1.000 atmx = 1.412 L (to 4 sig. fig.)

**Example #5:** When heated to high temperatures, silver oxide (Ag_{2}O) decomposes to form solid silver and oxygen gas. Calculate the volume of oxygen produced at STP by the decomposition of 7.44 g of Ag_{2}O.

**Example #6:** 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 +^{3}⁄_{2}O_{2}---> Fe_{2}O_{3}The 1.5 to 1 molar ratio (I'll use 3 to 2 just below. Just to be different.) between O

_{2}and Fe_{2}O_{3}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 O_{2} to Fe_{2}O_{3} molar ratio referenced above, determine moles of Fe_{2}O_{3} 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 #7:** 1.10 g of sodium superoxide, NaO_{2}, was reacted with excess water. Calculate the volume of oxygen gas measured at RTP produced in the reaction.

**Solution:**

1) Write the chemical reaction:

2NaO_{2}+ H_{2}O ---> 2NaOH +^{3}⁄_{2}O_{2}I have deliberately left it balanced with a fractional coefficient.

2) Determine moles of NaO_{2} that reacted:

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

3) NaO_{2} and O_{2} 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 O_{2} produced:

4 is to 3 as 0.0200 mol is to xx = 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 #8:** When 4.73 g of a solid was heated, the residue weighed 4.10 g and 320. cm^{3} 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 minus 4.10 g = 0.63 gYay for the Law of Conservation of Mass!

2) Determine the moles of gas produced:

PV = nRT(____) (0.320 L) = (n) (0.08206 L atm / mok 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 / mok 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 thisn = 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 #9:** 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 ---> ZnCl_{2}+ H_{2}The key molar ratio is between Zn and H

_{2}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 H_{2}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 H_{2} gas at SATP:

PV = nRT(100.0 kPa / 101.325 kPa/atm) (V) = (0.530743 mol) (0.08206 L atm mol¯

^{1}K¯^{1}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.

5) I saw this question answered on Yahoo Answers and the following was the answer:

(34.7 g Zn) / (65.38 g Zn/mol) x (1 mol H_{2}/ 1 mol Zn) x (24.8 L/mol) = 13.2 L H_{2}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:

P _{1}V_{1}P _{2}V_{2}––––– = ––––– T _{1}T _{2}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 Lto three sig figs, this is 24.8 L for the molar volume at STAP

**Example #10:** Solid lithium hydroxide is used to "scrub" CO_{2} from the air in spacecraft and submarines; it reacts with the CO_{2} to produce lithium carbonate and water. What volume in liters of CO_{2} 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 + CO_{2}---> Li_{2}CO_{3}+ H_{2}OThe molar ratio of importance is the 2:1 ratio between LiOH and CO

_{2}

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 CO_{2} that react:

2 is to 1 as 19.96 ml is to xx = 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 Problem:** Sodium reacts with water to form sodium hydroxide and hydrogen gas according to the equation:

2Na(s) + 2H_{2}O(ℓ) ---> 2NaOH(aq) + H_{2}(g)

(a) If 90.0 grams of sodium is dropped into 80.0 g of water, how many liters of hydrogen at STP would be produced?

(b) Which reactant is in excess and how much of it is left over?

**Solution:**

1) Let us determine how much H_{2} would be produced by 90.0 g of Na (assuming sufficient water):

90.0 g / 22.99 g/mol = 3.91474 molFor every two moles of Na, one mole of H

_{2}is produced.1.95737 mol of H

_{2}is produced

2) Let us determine how much H_{2} would be produced by 80.0 g of water (assuming sufficient sodium):

80.0 g / 18.015 g/mol = 4.44074 molFor every two moles of H

_{2}O, one mole of H_{2}is produced.2.22037 mole of H

_{2}is produced

3) Na runs out first. It is the limiting reagent. Water is the excess reagent. How much of it is left?

Note the 1:1 molar ratio between sodium and water.That means that 3.91474 mol of water was used up while the 3.91474 mol of Na was being used up completely.

4) The amount of water that remains is this (the answer to (b)):

4.44074 mol - 3.91474 mol = 0.526 mol(0.526 mol) (18.015 g/mol) = 9.48 g (to three sig figs)

4) How many liters of H_{2} is produced (the answer to (a))?

PV = nRT(1.00 atm) (V) = (1.95737 mol) (0.08206 L atm mol¯

^{1}K¯^{1}) (273 K)V = 43.85 L (rounds to 43.8 for three sig figs)

5) Since the conditions are at STP, the molar volume could be used:

(1.95737 mol) (22.414 L/mol) = 43.87 L (rounds to 43.9 L for 3 sig figs)