Stoichiometry
Mass-Volume Problems
#11 - 25

Probs #1-10

Ten Examples

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Comment: many stoichiometry problems involving volume will stipulate conditions of STP. This means that, if a volume is asked for as an answer, you can get to it by using: (a) PV = nRT or (b) molar volume. I will use one or the other, sometimes both, as the mood struck me when I wrote the solutions.


Problem #11: What volume of HCl gas will be produced at 25.0 °C and 2.75 atm from 5.00 g of H2 in the following reaction?

H2 + Cl2 ---> 2HCl

Solution:

5.00 g / 2.016 g/mol = 2.48016 mol of H2

H2 to HCl molar ratio is 1:2

(2.48016 mol) (2) = 4.96032 mol of HCl produced

PV = nRT

(2.75 atm) (V) = (4.96032 mol) (0.08206 L atm / mol K) (298 K)

V = 44.1 L (to three sig figs)


Problem #12: Phosphorus burns in oxygen gas to produce phosphorus(V) oxide (P4O10) according to the equation:

P4(s) + 5O2(g) ---> P4O10(g)

(a) If 2.50 grams of phosphorus is ignited together with 750. mL of oxygen at STP, how many grams of P4O10 are formed?
(b) Which reactant is in excess and how much of it is left over?

Solution:

1) Let us determine how much P4O10 would be produced by 2.50 g of P4 (assuming sufficient oxygen. In other words, ignore the 750. mL of oxygen):

2.50 g / 123.896 g/mol = 0.020178 mol

P4 and P4O10 are in a 1:1 molar ratio.

This means 0.020178 mol of P4O10 would be produced.

2) Let us determine how much P4O10 would be produced by 750. mL of O2 (assuming sufficient phosphorus. In other words, ignore the 2.50 g of phosphorus.):

PV = nRT

(1.00 L) (0.750 L) = (n) (0.08206 L atm mol¯11) (273 K)

n = 0.03347858 mol of O2(g) present

O2 and P4O10 are in a 5:1 molar ratio

This means 0.006695716 mol of P4O10 would be produced.

3) Oxygen gas is the limiting reagent. Phosphorous is the reagent in excess. How much of it remains?

0.006695716 mol of P4O10 is produced.

P4 and P4O10 are in a 1:1 molar ratio.

0.006695716 mol of P4 is consumed.

(0.006695716 mol) (123.896 g/mol) = 0.82957 g

2.50 − 0.83 = 1.67 g remaining


Problem #13: Magnesium burns in oxygen to produce magnesium oxide.

(a) If 1.00 gram of magnesium is ignited together with 0.500 liter of oxygen at STP, how many grams of magnesium oxide are produced?
(b) What is the name and amount of the reactant in excess?

Solution:

1) We need a balanced chemical equation:

Mg(s) + 12O2(g) ---> MgO(s)

I deliberately used a fractional coefficient.

2) I will use a different method than above to get to the limiting reagent:

(a) Determine moles of each reactant:
1.00 g / 24.305 g/mol = 0.0411438 mol

(1.00 atm) (0.500 L) = (n) (0.08206 L atm / mol K) (273 K); n = 0.022319 mol

(b) divide by coefficient in balanced equation:

0.0411438 / 1 = 0.0411438 <--- Mg is the lower amount

0.022319 / 0.5 = 0.044638

(c) Magnesium is the limiting reagent. The oxygen gas is the reagent in excess.

4) Determine the amount of MgO produced:

Mg and MgO are in a 1:1 molar ratio.

Therefore, 0.0411438 mol of MgO is produced.

(0.0411438 mol) (40.304 g/mol) = 1.66 g (to three sig figs)

5) Determine the volume of oxygen that remains:

Mg and O2 react in a 1 : 0.5 molar ratio

Therefore, 0.0205719 mol of O2 was consumed.

PV = nRT

(1.00 atm) (V) = (0.0205719 mol) (0.08206 L atm / mol K) (273 K)

V = 0.46086 L

0.500 L − 0.46086 L = 0.039 L = 39 mL


Problem #14: MnO2(s) + 4HCl(aq) ---> MnCl2(aq) + 2H2O(ℓ) + Cl2(g)

(a) How many grams of HCl are required to produce 224 liters of Cl2 at 27.0 °C and 750. mm Hg?
(b) How many liters of chlorine (Cl2) at 20.0 °C and 765 mmHg will be produced from 1.00 x 103 g of manganese dioxide (MnO2)?

Solution to (a):

1) Determine moles of Cl2 that get produced:

PV = nRT

(750. mmHg / 760. mmHg/atm) (224 L) = (n) (0.08206 L atm / mol K) (300. K)

n = 8.97931 mol

2) Determine moles, then grams of HCl required

The HCl to Cl2 molar ratio is 4 to 1.

(8.97931 mol) (4) = 35.91724 mol of HCl required

(35.91724 mol) (36.4609 g/mol) = 1310 g

Solution to (b):

1) Determine moles of MnO2:

1.00 x 103 g / 86.936 g/mol = 11.5027 mol of MnO2

2) Determine moles of Cl2 produced:

MnO2 and Cl2 are in a 1:1 molar ratio

11.5027 mol of Cl2 is produced

3) Determine volume of Cl2:

PV = nRT

(765 mmHg / 760. mmHg/atm) (V) = (11.5027 mol) (0.08206 L atm / mol K) (293 K)

V = 275 L (to three sig figs)


Problem #15: Turpentine (C10H16) burns in chlorine (Cl2) to produce carbon (C) and hydrogen chloride (HCl) according to the equation:

C10H16(ℓ) + 8Cl2(g) ---> 10C(s) + 16HCl(g)

(a) How many liters of HCl at 100.0 °C and 740. mmHg are produced from 150.0 g of C10H16?
(b) If 200.0 L of Cl2 at 25.0 °C and 800. mmHg reacts with sufficient C10H16, how many grams of C are produced?

Solution to (a):

1) Moles of turpentine:

150.0 g / 136.2364 g/mol = 1.10103 mol

2) Moles of HCl:

The turpentine to HCl molar ratio is 1:16.

(1.10103 mol) (16) = 17.61648 mol

3) Volume of HCl:

PV = nRT

(740. mmHg / 760. mmHg/atm) (V) = (17.61648 mol) (0.08206 L atm / mol K) (373 K)

V = 30.6 L (to 3 sig figs)

Solution to (b):

1) Determine moles of Cl2:

PV = nRT

(800. mmHg / 760. mmHg / atm) (200.0 L) = (n) (0.08206 L atm / mol K) (298 K)

n = 8.609117 mol

2) Determine moles, then grams of C:

Cl2 and C are in a 4:5 molar ratio

4 is to 5 as 8.609117 is to x

x = 10.76139625 mol

(10.76139625 mol) (12.011 g/mol) = 129 g


Problem #16: 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)


Problem #17: One of the steps in the production of iron utilizes the following chemical reaction:

3CO(g) + Fe2O3(s) ---> 2Fe(s) + 3CO2(g)

(a) What mass of Fe2O3 would react with 200.0 liters of CO at STP?
(b) What volume of carbon dioxide (CO2) at STP is produced from 100.0 grams of Fe2O3?
(c) What mass of iron (Fe) is produced when 300. mL of CO2 is produced at STP?

Solution to (a):

1) Determine moles of CO:

200.0 L / 22.414 L/mol = 8.923 mol

Note use of molar volume in this problem (and the one above) rather than PV = nRT. Since we are at STP, we can use molar volume. Any other conditions require the use of PV = nRT.

2) Determine moles, then mass, of Fe2O3:

CO and Fe2O3 react in a 3:1 molar ratio.

8.923 mol is to x as 3 is to 1

x = 2.97433 mol

(2.97433 mol) (159.687 g/mol) = 475 g

Solution to (b):

1) Moles of Fe2O3 consumed:

100.0 g / 159.687 g/mol = 0.626225 mol

2) Moles of CO2 produced:

Fe2O3 and CO2 are in a 1:3 molar ratio.

(0.626225 mol) (3) = 1.878675 mol

3) Determine volume of CO2 produced:

PV = nRT

(1.00 atm) (V) = (1.878675 mol) (0.08206 L atm / mol K) (273 K)

V = 42.1 L

Solution to (c):

1) Moles of CO2. Use molar volume:

0.300 L / 22.414 L/mol = 0.0133845 mol

2) Moles of Fe:

Iron and carbon dioxide are in a 2:3 molar ratio

2 is to 3 as x is to 0.0133845

x = 0.008923 mol of Fe

3) Mass of Fe:

(0.008923 mol) (55.845 g/mol) = 0.498 g

Problem #18: How many liters of O2 gas at STP will be produced by 54.0 g of HgO in the following reaction?

2HgO ---> 2Hg + O2

Solution:

1) Determine moles of HgO:

54.0 g / 216.589 g/mol = 0.24932 mol

2) Determine moles of oxygen gas produced:

HgO to O2 molar ratio is 2 to 1

0.24932 mol / 2 = 0.12466 mol of O2 produced

3) Since we are at STP, use molar volume:

(0.12466 mol) (22.414 L/mol) = 2.79 L (to three sig figs)

Problem #19: How many liters of H2 gas at STP will be produced by 0.400 mole of Ca in the following reaction?

3Ca + 2H3PO4 ---> Ca3(PO4)2 + 3H2

Solution:

1) There is a one-to-one molar ratio between Ca and H2. That means 0.400 mol of H2 will be produced.

2) Use molar volume:

(22.414 L/mol) (0.400 mol) = 8.96 L (to three sig figs)

Remember, using molar volume only works at STP.

Use of PV = nRT to get 8.96 L is left to the student.


Problem #20: Given the balanced equation:

CH4(g) + 2O2(g) ---> CO2(g) + 2H2O(g)

(a) How many moles of carbon dioxide (CO2) are formed when 40.0 liters of oxygen (O2) at STP is consumed?
(b) How many liters of methane (at RTP) are needed to form 200. moles of water?

Solution to (a):

1) The oxygen to carbon dioxide molar ratio is two-to-one.

2   40.0 L
–––––––  =  –––––––
1   x

x = 20.0 L of CO2 are formed

Comment: at conditions of equal temperature and pressure, a ratio of moles and a ratio of volumes are equivalent statements.

2) Determine moles:

PV = nRT

(1.00 atm) (20.0 L) = (n) (0.08206 L atm / mol K) (273 K)

n = 0.893 mol

The solution using molar volume is left to the student.

Solution to (b):

1) Determine moles of CH4:

the methane:water molar ratio is 1:2

1 is to 2 as x is to 200.

x = 100. mol of methane required

2) Determine volume at RTP:

PV = nRT

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

V = 2404.358

to three sig figs, 2.40 x 103 L

Comment: 2400 would be two sig figs and 2400. would be 4.


Problem #21: Given the balanced equation:

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

If 0.840 g of ammonium dichromate is used, and 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 and (b) what are the partial pressures of N2 and H2O?

Solution to (a):

1) Determine moles of ammonium dichromate:

0.840 g / 252.0622 g/mol = 0.0033325 mol

2) Determine moles of gas produced:

One mole of ammonium dichromate produces five moles of gas (one of nitrogen and 4 of water).

1   0.0033325 mol
–––  =  –––––––––––––
5   x

x = 0.0166625 mol

3) Use PV = nRT:

(P) (13.6 L) = (0.0166625 mol) (0.08206 L atm / mol K) (288 K)

P = 33.3 atm (to three sig figs)

Solution to (b):

1) Determine the mole fractions for N2 and H2O:

N2 ---> 1/5 = 0.200
H2O ---> 4/5 = 0.800

2) Determine partial pressures:

N2 ---> (0.200) (33.26565 atm) = 6.7 atm (actually determined by subtraction)
H2O ---> (0.800) (33.3 atm) = 26.6 atm (to three sig figs)

Problem #22: Nitroglycerin (227.1 g/mol) decomposes according to the reaction below.

4C3H5N3O9(ℓ) ---> 6N2(g) + 12CO2(g) + 10H2O(g) + O2(g)

What total volume of gases are produced at 33.0 °C and 0.968 atm by the decomposition of 2.00 g nitroglycerin?

Solution:

1) Let's get the moles of nitroglycerin:

2.00 g / 227.1 g/mol = 0.0088067 mol

2) Four moles of nitroglycerin produce 29 moles of gas:

4   0.0088067 mol
–––  =  ––––––––––––
29   x

x = 0.063848575 mol

3) Let's use PV = nRT:

(0.968 atm) (V) = (0.063848575 mol) (0.08206 L atm mol-1 K-1) (306 K)

V = 1.66 L


Problem #23: When coal is burned, the sulfur present in coal is converted to sulfur dioxide (SO2):

S8(s) + 8O2(g) ---> 8SO2(g)

If 3.25 kg of S8 are reacted with oxygen, calculate the volume of SO2 gas (in mL) formed at 30.9 °C and 1.09 atm.

Solution:

1) Determine moles of sulfur:

3250 g / 256.52 g/mol = 12.6696 mol

2) The S8 to SO2 molar ratio is 1:8.

1   12.6696 mol
–––  =  ––––––––––
8   x

x = 101.3568 mol

3) Use PV = nRT

(1.09 atm) (V) = (101.3568 mol) (0.08206 L atm mol-1 K-1) (303.9 K)

V = 2320 L (to three sig figs)

4) The question asked for mL:

2.32 x 106 mL

Problem #24: What volume (at STP) of chlorine gas is produced by the decomposition of 25.0 grams of AlCl3?

Solution:

1) Write the balanced chemical equation:

2AlCl3(s) ---> 2Al(s) + 3Cl2(g)

2) Determine moles of AlCl3:

25.0 g / 133.341 g/mol = 0.18749 mol

3) Determine moles of Cl2 produced:

The molar ratio between AlCl3 and Cl2 is 2:3

2   0.18749 mol
–––––––  =  ––––––––––
3   x

x = 0.281235 mol (of chlorine gas produced)

4) Since we are at STP, let us use molar volume:

(22.414 L/mol) (0.281235 mol) = 6.30 L

Problem #25: What mass of phosphorus would be required to react with 110. L of oxygen gas measured at 98.0 °C and 92.80 kPa to produce diphosphorus pentoxide?

Solution:

1) Write the balanced chemical equation:

P4 + 5O2 ---> 2P2O5

2) Use PV = nRT to determine moles of oxygen gas:

(92.80 kPa) (110. L) = (n) (8.31447 L kPa / mol K) (371 K)

n = 3.26994 mol

I obtained the value for R here. Second table, the column headed 'liter' and the row labeled 'kPa.'

3) Determine moles of phosphorus required:

The P4 to O2 molar ratio is 1 to 5.

1   x
–––––––  =  ––––––––––
5   3.26994 mol

x = 0.653988 mol

4) Determine mass of phosphorus required:

(0.653988 mol) (123.896 g/mol) = 81.0 g

Problem #26: A reaction between hydrochloric acid and potassium sulfide was performed. 196.3 mL of a gas product was collected over water at 20.0 °C and a total pressure of 753.70 torr. How many grams of potassium sulfide reacted? (water's vapor pressure at 20.0 °C is 17.54 mmHg)

Solution:

1) Write the chemical reaction:

2HCl(g) + K2S(s) ---> 2KCl(s) + H2S(g)

2) Using Dalton's Law, determine the pressure of the dry gas:

Ptotal = PH2S + PH2O

753.70 torr = PH2S + 17.54 torr

PH2S = 736.16 torr

3) Using the Ideal Gas Law, determine the moles of H2S produced:

PV = nRT

(736.16 torr / 760.0 torr/atm) (0.1963 L) = (n) (0.08206 L atm / mol K) (293 K)

n = 0.00790824 mol

4) Using the molar ratio between K2S and H2S, determine the moles of K2S that reacted:

1   0.00790824 mol
–––  =  –––––––––––––
1   x

x = 0.00790824 mol

5) Using the molar mass of K2S, determine grams of K2S that reacted:

(0.00790824 mol) (110.262 g/mol) = 0.872 g (to three sig figs)

Problem #27: 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)

Problem #28: If you start with 145.0 g Fe3O4, what is the percent yield of the reaction if 38.0 liters of CO2 were produced?

Fe3O4 + 4CO ---> 3Fe + 4CO2

Solution:

1) Calculate the moles of CO2 produced. Since no temperature or pressure is given, we will assume conditions of STP.

38.0 L / 22.414 L/mol = 1.69537 mol (I will keep some guard digits)

The assumption of STP allowed me to use molar volume. If I had assumed non-STP conditions, I would have needed to use PV = nRT to calculate the number of moles.

2) Determine moles of Fe3O4 consumed when 1.69537 mol of CO2 is produced. The Fe3O4 to CO2 molar ratio is 1:4.

1   x
–––––––  =  –––––––
4   1.69537 mol

x = 0.4238425 mol of Fe3O4

3) Determine mass of Fe3O4:

(0.4238425 mol) (231.531 g/mol) = 98.13268 g (I'll round off to final sig figs at the percent yield)

4) Determine percent yield:

98.13268 g/mol / 145.0 g = 67.68% yield

Problem #29: What mass of sodium bicarbonate would have to be added to an excess of dilute HCl(aq) in order to generate 1.50 L of carbon dioxide at RTP?

Solution:

Comment: carbon dioxide is soluble in water to the extent of 1.45 g/L at RTP. We will completely ignore the solubility of CO2 in water and treat it as being insoluble in water.

1) Let us write the chemical equation:

NaHCO3(s) + HCl(aq) ---> NaCl(aq) + CO2(g) + H2O(ℓ)

The equation is balanced as written.

2) The key will be the molar ratio between NaHCO3 and CO2. It is a 1:1 ratio.

3) RTP stands for 'room temperature and pressure.' It is not a standard measurement in chemistry. The most common set of values used is 20.0 °C and 1.00 atm. I will use those values.

4) Determine moles of CO2 in 1.50 L at RTP:

PV = nRT

(1.00 atm) (1.50 L) = (n) (0.08206 L atm / mol K) (298 K)

n = 0.06133996 mol

5) The 1:1 molar ratio means this:

0.0613399 mol of CO2 was produced by 0.06133996 mol of NaHCO3

6) Determine mass of sodium bicarbonate that would have been needed to have been added:

(0.0613399 mol) (84.0059 g/mol) = 5.15 g (to three sig figs)

Bonus Problem: Assuming air is composed of 0.0407% carbon by volume, what is the volume of air required to produce 1.00 kg of glucose at NTP?

6CO2 + 6H2O ---> C6H12O6 + 6O2

Solution:

1) Determine moles of glucose:

1000 g / 180.1548 g/mol = 5.55078 mol

2) Determine moles of CO2 required:

The molar ratio between CO2 and glucose is 6:1

6 is to 1 as x is to 5.55078

x = 33.30468 mol

3) Determine volume of CO2 at NTP:

PV = nRT

(1.00 atm) (V) = (33.30468 mol) (0.08206 L atm / mol K) (293 K)

V = 800.764 L

I used values for NTP found here:

"NIST uses a temperature of 20 °C (293.15 K, 68 °F) and an absolute pressure of 1 atm (14.696 psi, 101.325 kPa). This standard is also called normal temperature and pressure (abbreviated as NTP)."

NTP is sometimes called RTP (room temperature and pressure).

4) Determine volume of air required to deliver 800.764 L of carbon dioxide:

800.764 L is to 0.0407% as x is to 100%

x = 1967479 L

It can also be set up this way:

0.0407% means a decimal portion of 0.000407

800.764 L is to 0.000407 as x is to 1

To three sig figs, 1970000 L (or, 1.97 x 106 L)


Probs #1-10

Ten Examples

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