Mass-Volume Problems

#11 - 25

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 H_{2} in the following reaction?

H_{2}+ Cl_{2}---> 2HCl

**Solution:**

5.00 g / 2.016 g/mol = 2.48016 mol of H_{2}H

_{2}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 (P_{4}O_{10}) according to the equation:

P_{4}(s) + 5O_{2}(g) ---> P_{4}O_{10}(g)

(a) If 2.50 grams of phosphorus is ignited together with 750. mL of oxygen at STP, how many grams of P_{4}O_{10} are formed?

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

**Solution:**

1) Let us determine how much P_{4}O_{10} would be produced by 2.50 g of P_{4} (assuming sufficient oxygen. In other words, ignore the 750. mL of oxygen):

2.50 g / 123.896 g/mol = 0.020178 molP

_{4}and P_{4}O_{10}are in a 1:1 molar ratio.This means 0.020178 mol of P

_{4}O_{10}would be produced.

2) Let us determine how much P_{4}O_{10} would be produced by 750. mL of O_{2} (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¯

^{1}K¯^{1}) (273 K)n = 0.03347858 mol of O

_{2}(g) presentO

_{2}and P_{4}O_{10}are in a 5:1 molar ratioThis means 0.006695716 mol of P

_{4}O_{10}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 P_{4}O_{10}is produced.P

_{4}and P_{4}O_{10}are in a 1:1 molar ratio.0.006695716 mol of P

_{4}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) +^{1}⁄_{2}O_{2}(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 amount0.022319 / 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 O_{2}react in a 1 : 0.5 molar ratioTherefore, 0.0205719 mol of O

_{2}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:** MnO_{2}(s) + 4HCl(aq) ---> MnCl_{2}(aq) + 2H_{2}O(ℓ) + Cl_{2}(g)

(a) How many grams of HCl are required to produce 224 liters of Cl_{2} at 27.0 °C and 750. mm Hg?

(b) How many liters of chlorine (Cl_{2}) at 20.0 °C and 765 mmHg will be produced from 1.00 x 10^{3} g of manganese dioxide (MnO_{2})?

**Solution to (a):**

1) Determine moles of Cl_{2} 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 Cl_{2}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 MnO_{2}:

1.00 x 10^{3}g / 86.936 g/mol = 11.5027 mol of MnO_{2}

2) Determine moles of Cl_{2} produced:

MnO_{2}and Cl_{2}are in a 1:1 molar ratio11.5027 mol of Cl

_{2}is produced

3) Determine volume of Cl_{2}:

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 (C_{10}H_{16}) burns in chlorine (Cl_{2}) to produce carbon (C) and hydrogen chloride (HCl) according to the equation:

C_{10}H_{16}(ℓ) + 8Cl_{2}(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 C_{10}H_{16}?

(b) If 200.0 L of Cl_{2} at 25.0 °C and 800. mmHg reacts with sufficient C_{10}H_{16}, 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 Cl_{2}:

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:

Cl_{2}and C are in a 4:5 molar ratio4 is to 5 as 8.609117 is to x

x = 10.76139625 mol

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

**Problem #16:** 12CO_{2}(g) + 11H_{2}O(ℓ) ---> C_{12}H_{22}O_{11}(s) + 12O_{2}(g)

(a) How many grams of sucrose (C_{12}H_{2}2O_{11}) are produced from 224 liters of carbon dioxide
(CO_{2}) at STP?

(b) How many liters of carbon dioxide (CO_{2}) 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 CO_{2}, measured at RTP?

**Solution to (a):**

1) Determine moles of CO_{2}:

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 CO_{2} required:

There is a 12:1 molar ratio between CO_{2}and sucrose.(6.625764 mol) (12) = 79.509168 mol

3) Determine volume of CO_{2}:

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 actually has two different sets of values:

(a) 20.0 °C and 1.00 atm

b) 25.0° and 1.00 atmI will use (a), the one used by the National Institute of Standards and Technology (NIST).

2) Determine moles of CO_{2}:

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 H_{2}O:

There is a 12:11 molar ratio between CO_{2}and H_{2}O.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) + Fe_{2}O_{3}(s) ---> 2Fe(s) + 3CO_{2}(g)

(a) What mass of Fe_{2}O_{3} would react with 200.0 liters of CO at STP?

(b) What volume of carbon dioxide (CO_{2}) at STP is produced from 100.0 grams of Fe_{2}O_{3}?

(c) What mass of iron (Fe) is produced when 300. mL of CO_{2} is produced at STP?

**Solution to (a):**

1) Determine moles of CO:

200.0 L / 22.414 L/mol = 8.923 molNote 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 Fe_{2}O_{3}:

CO and Fe_{2}O_{3}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 Fe_{2}O_{3} consumed:

100.0 g / 159.687 g/mol = 0.626225 mol

2) Moles of CO_{2} produced:

Fe_{2}O_{3}and CO_{2}are in a 1:3 molar ratio.(0.626225 mol) (3) = 1.878675 mol

3) Determine volume of CO_{2} 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 CO_{2}. 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 ratio2 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 O_{2} gas at STP will be produced by 54.0 g of HgO in the following reaction?

2HgO ---> 2Hg + O_{2}

**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 O_{2}molar ratio is 2 to 10.24932 mol / 2 = 0.12466 mol of O

_{2}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 H_{2} gas at STP will be produced by 0.400 mole of Ca in the following reaction?

3Ca + 2H_{3}PO_{4}---> Ca_{3}(PO_{4})_{2}+ 3H_{2}

**Solution:**

1) There is a one-to-one molar ratio between Ca and H_{2}. That means 0.400 mol of H_{2} 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:

CH_{4}(g) + 2O_{2}(g) ---> CO_{2}(g) + 2H_{2}O(g)

(a) How many moles of carbon dioxide (CO_{2}) are formed when 40.0 liters of oxygen (O_{2}) 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 CO

_{2}are formedComment: 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 CH_{4}:

the methane:water molar ratio is 1:21 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 10

^{3}LComment: 2400 would be two sig figs and 2400. would be 4.

**Problem #21:** Given the balanced equation:

(NH_{4})_{2}Cr_{2}O_{7}(s) ---> N_{2}(g) + 4H_{2}O(g) + Cr_{2}O_{3}(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 N_{2} and H_{2}O?

**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 N_{2} and H_{2}O:

N_{2}---> 1/5 = 0.200

H_{2}O ---> 4/5 = 0.800

2) Determine partial pressures:

N_{2}---> (0.200) (33.26565 atm) = 6.7 atm (actually determined by subtraction)

H_{2}O ---> (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.

4C_{3}H_{5}N_{3}O_{9}(ℓ) ---> 6N_{2}(g) + 12CO_{2}(g) + 10H_{2}O(g) + O_{2}(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 (SO_{2}):

S_{8}(s) + 8O_{2}(g) ---> 8SO_{2}(g)

If 3.25 kg of S_{8} are reacted with oxygen, calculate the volume of SO_{2} 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 S_{8} to SO_{2} 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 10^{6}mL

**Problem #24:** What volume (at STP) of chlorine gas is produced by the decomposition of 25.0 grams of AlCl_{3}?

**Solution:**

1) Write the balanced chemical equation:

2AlCl_{3}(s) ---> 2Al(s) + 3Cl_{2}(g)

2) Determine moles of AlCl_{3}:

25.0 g / 133.341 g/mol = 0.18749 mol

3) Determine moles of Cl_{2} produced:

The molar ratio between AlCl_{3}and Cl_{2}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:

P_{4}+ 5O_{2}---> 2P_{2}O_{5}

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 P_{4}to O_{2}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

**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?

6CO_{2}+ 6H_{2}O ---> C_{6}H_{12}O_{6}+ 6O_{2}

**Solution:**

1) Determine moles of glucose:

1000 g / 180.1548 g/mol = 5.55078 mol

2) Determine moles of CO_{2} required:

The molar ratio between CO_{2}and glucose is 6:16 is to 1 as x is to 5.55078

x = 33.30468 mol

3) Determine volume of CO_{2} 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 10

^{6}L)