### StoichiometryMass-Volume Problems#11 - 25

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:

1) Calculations using mass data:

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

2) Volumetric calculation:

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: 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 #13: 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 #14: 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 #15: 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 #16: 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 #17: 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 #18: 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 #19: 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 #20: 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)

Problem #21: Urea can be made by the reaction of carbon dioxide and ammonia:

CO2(g) + 2NH3(g) ---> CO(NH2)2(s) + H2O(g)

Assuming the reaction goes to completion:

(a) What volume of CO2 at 200. atm and 341°C is needed to produce 4.30 kg of urea?
(b) What volume of NH3 at the same conditions is needed to produce 4.30 kg of urea?

Solution:

1) Determine moles of urea produced:

4300 g / 60.0556 g/mol = 71.6 mol

2) Determine moles of CO2 required:

Based on 1:1 molar ratio between CO2 and urea, we determine 71.6 mol of CO2 are required.

3) Determine the volume occupied by the CO2 at the given T and P:

PV = nRT

(200. atm) (V) = (71.6 mol) (0.08206 L atm / mol K) (614 K)

V = 18.0 L (to three sig figs)

4) Based on the 2:1 molar ratio between ammonia and carbon dioxide, 36.0 L of NH3 is required.

Problem #22: How many molecules of sulfur trioxide would react with 68.9 L of water at STP to produce sulfuric acid?

SO3 + H2O ---> H2SO4

Solution:

1) Determine moles of water present in the 68.9 L. I'll use molar volume.

68.9 L / 22.414 L/mol = 3.07397 mol

2) Determine moles of SO3 that react with 3.07397 mol of water.

There is a 1:1 molar ratio between the reactants.

3.07397 mol of SO3 is consumed.

3) Use Avogadro's Number to determine number of molecules.

(3.07397 mol) (6.022 x 1023 molecules/mol) = 1.85 x 1024 molecules of SO3

Problem #23: CaC2 reacts violently with water to make calcium hydroxide and acetylene gas (C2H2) as shown below:

CaC2(s) + H2O(ℓ) ---> Ca(OH)2(aq) + C2H2(g)

16.0 L of acetylene gas is collected over water at a temperature of 49.0 °C and a pressure of 2371 mmHg. Given that the vapor pressure of water at 49.0 °C is 119 mmHg, how many grams of CaC2 reacted?

Solution:

1) Determine the pressure of the dry acetylene gas:

x + 119 mmHg = 2371 mmHg

x = 2252 mmHg

2) Determine moles of dry gas:

(2252 mmHg / 760.0 mmHg/atm) (16.0 L) = (y) (0.08206 L atm/mol K) (322 K)

y = 1.79427 mol

3) The molar ratio between calcium carbide and acetylene gas is 1:1. Therefore:

1.79427 mol of CaC2 reacted

4) Determine mass of calcium carbide that reacted:

(1.79427 mol) (64.099 g/mol ) = 115 g

Problem #24: 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 #25: 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.

Bonus Problem #1: 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)

Bonus Problem #2: 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 / 145.0 g = 67.7% yield (to three sig figs)