Gas Stoichiometry

Gas Stoichiometry

• We have looked at stoichiometry: 1) using masses & molar masses, & 2) concentrations.

• We can use stoichiometry for gas reactions.

• As before, we need to consider mole ratios when examining reactions quantitatively.

molar mass of x mole ratio from balanced equation molar mass of y grams (x)  moles (x)  moles (y)  grams (y) PV = nRT P, V, T (x) P, V, T (y) • At times you will be able to use 22.4 L/mol at STP as a shortcut.

Sample problem 1

CH 4 CH 4 burns in O 2 , producing CO 2 and H 2 O(g). A 1.22 L cylinder, at 15 °C, registers a pressure of 328 kPa.

a) What volume of O 2 at 100 kPa and 25 o C will be required to react completely with all of the CH 4 ?

First: CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g) PV = nRT P = 328 kPa, V = 1.22 L, T = 288 K (3.24 atm)(1.22 L) (.0821 atm •L/K•mol)(288 K)

# mol O 2 = 0.167 mol CH

PV = nRT

4 x

= n = 0.167 mol

2 mol O 2 1 mol CH 4 = 0.334 mol

P= 100 kPa, n= 0.334 mol, T= 298 K (0.334 mol)(.0821 atm •L/K•mol )(298 K) =V = 8.28 L (.987 atm)

Sample problem 1 continued

CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g) b) How many grams of H 2 O(g) are produced?

# g H 2 O=0.167 mol CH 4 x 2 mol H 2 O 1 mol CH 4 x 18.02 g H 2 O 1 mol H 2 O = 6.02 g H 2 O

c) What volume of CO 2 (at STP) is produced if only 2.15 g of the CH 4

# mol CO 2 =2.15 g CH 4 x

(1.00 atm) was burned?

1 mol CH 4 16.05 g CH 4 x

PV = nRT (0.134 mol)(.0821 atm •L/K•mol )(273 K)

1 mol CO 2 1 mol CH 4 = 0.134

mol CO 2

P = 1.00 atm, n = 0.134 mol, T = 273 K = V = 3.00 L CO 2 or # L = 0.134 mol x 22.4 L/mol = 3.00 L

Sample problem 2

Ammonia (NH 3 ) gas can be synthesized from nitrogen gas + hydrogen gas. What volume of ammonia at 450 kPa and 80 °C can be obtained from the complete reaction of 7.5 kg hydrogen?

First we need a balanced equation: N 2 (g) + 3H 2 (g)  2NH 3 (g)

# mol NH 3 =7500 g H 2 x 1 mol H 2 2.02 g H 2 x 2 mol NH 3 3 mol H 2

PV = nRT

= 2475 mol

P = 450 kPa, n = 2475 mol, T = 353 K (2475 mol)(.0821)(353 K) (4.44 atm) = V = 16 135 L NH 3

Sample problem 3

Hydrogen gas (and NaOH) is produced when sodium metal is added to water. What mass of Na is needed to produce 20.0 L of H 2 at STP?

First we need a balanced equation: 2Na(s) + 2H 2 O(l)  H 2 (g) + 2NaOH(aq) PV = nRT P= 1.00 atm, V= 20.0 L, T= 273 K (1.00 atm)(20.0 L) = n = 0.893 mol H (.0821 atm •L/K•mol )(273 K) or # mol = 20.0 L x 1 mol / 22.4 L = 0.893 mol 2

# g Na= 0.893 mol H 2 x 2 mol Na 1 mol H 2 x 22.99 g Na 1 mol Na = 41.1 g Na

Assignment

1. What volume of oxygen at STP is needed to completely burn 15 g of methanol (CH 3 OH) in a fondue burner? (CO 2 + H 2 O are products) 2. When sodium chloride is heated to 800 °C it can be electrolytically decomposed into Na metal & chlorine (Cl 2 ) gas. What volume of chlorine gas is produced (at 800 °C and 100 kPa) if 105 g of Na is also produced?

3. What mass of propane (C 3 H 8 ) can be burned using 100 L of air at SATP? Note: 1) air is 20% O 2 , so 100 L of air holds 20 L O 2 , 2) CO 2 and H 2 O are the products of this reaction.

4. A 5.0 L tank holds 13 atm of propane (C 3 H 8 ) at 10 °C. What volume of O 2 at 10 °C & 103 kPa will be required to react with all of the propane?

5. Nitroglycerin explodes according to: 4 C 3 H 5 (NO 3 ) 3 (l)  12 CO 2 (g) + 6 N 2 (g) + 10 H 2 O(g) + O 2 (g) a) Calculate the volume, at STP, of each product formed by the reaction of 100 g of C 3 H 5 (NO 3 ) 3 .

b) 200 g of C 3 H 5 (NO 3 ) 3 is ignited (and completely decomposes) in an otherwise empty 50 L gas cylinder. What will the pressure in the cylinder be if the temperature stabilizes at 220 °C?

Answers

1. 3O 2 (g) + 2CH 3 OH(l)  2CO 2 (g) + 4H 2 O(g)

# L O 2 = 15 g CH 3 x 3 OH 32.05 g CH 3 OH x 3 mol O 2 2 mol CH 3 OH x 22.4 L O 2 1 mol O 2 = 15.7 L O 2

2. 2NaCl(l)  2Na(l) + Cl 2 (g)

# mol Cl 2 = 105 g Na 1 mol Na 22.99 g Na x 1 mol Cl 2 2 mol Na = 2.284 mol Cl 2

PV = nRT P = .987 atm, n = 2.284 mol, T = 1073 K (2.284 mol)(.0821)(1073 K) (.987 atm) = V = 204 L Cl 2

3. C 3 H 8 (g) + 5O 2 (g)  3CO 2 (g) + 4H 2 O(g)

# g C 3 H 8 = 20 L O 2 x 1 mol O 2 24.8 L O 2 x 1 mol C 3 5 mol O H 2 8 x 44.11 g C 3 H 8 1 mol C 3 H 8 = 7.1 g C 3 H 8

4. C 3 H 8 (g) + 5O 2 (g)  3CO 2 (g) + 4H 2 O(g) PV = nRT n = (13.0 atm)(5.0 L) (.0821)(283 K) = 2.8 mol C 3 H

# mol O

PV = nRT

2 = 2.8 mol C 3 H 8 x 5 mol O 2 1 mol C 3 H 8 = 14 mol O

P = 1.02 atm, n = 14 mol, T = 283 K

2

8 (14 mol)(.0821)(283 K) = V = 320 L O (1.02 atm) 2

5.

# mol C 3 H 5 (NO 3 ) 3 = 100 g C 3 H 5 (NO 3 ) 3 x 1 mol C 3 H 5 (NO 3 ) 3 227.11 g C 3 H 5 (NO 3 ) 3 = 0.4403 mol # L CO 2 = 0.4403 mol 3 5 (NO 3 ) 3 x 12 mol CO 2 4 mol C 3 H 5 (NO 3 ) 3 x 22.4 L 1 mol = 29.6 L CO 2 # L N 2 = 0.4403 mol 3 5 (NO 3 ) 3 x 6 mol N 2 4 mol C 3 H 5 (NO 3 ) 3 x 22.4 L 1 mol = 14.8 L N 2 # L H 2 O= 0.4403 mol 3 5 (NO 3 ) 3 x 10 mol H 2 O 4 mol C 3 H 5 (NO 3 ) 3 x 22.4 L 1 mol = 24.7 L H 2 O # L O 2 = 0.4403 mol 3 5 (NO 3 ) 3 x 1 mol O 2 4 mol C 3 H 5 (NO 3 ) 3 x 22.4 L 1 mol = 2.47 L O 2

5.

# mol C 3 H 5 (NO 3 ) 3 = 200 g C 3 H 5 (NO 3 ) 3 x 1 mol C 3 H 5 (NO 3 ) 3 227.11 g C 3 H 5 (NO 3 ) 3 # mol all gases= 0.8806 mol C 3 H 5 (NO 3 ) 3 x 29 mol gases 4 mol C 3 H 5 (NO 3 ) 3 = 0.8806 mol = 6.385 mol all gases

PV = nRT V = 50 L, n = 6.385 mol, T = 493 K (6.385 mol)(.0821)(493 K) (50 L) = P = 5.16 atm For more lessons, visit www.chalkbored.com