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. At times you will be able to use 22.4 L/mol at STP and 24.8 L/mol at SATP as shortcuts. grams (x)  moles (x)  moles (y)  grams (y) molar mass of y mole ratio from balanced equation molar mass of x P, V, T (x)  P, V, T (y)  PV = nRT

Section 13.3 Gas Stoichiometry Determine volume ratios for gaseous reactants and products by using coefficients from chemical equations. coefficient: the number written in front of a reactant or product in a chemical equation, which tells the smallest number of particles of the substance involved in the reaction Apply gas laws to calculate amounts of gaseous reactants and products in a chemical reaction. When gases react, the coefficients in the balanced chemical equation represent both molar amounts and relative volumes.

Stoichiometry of Reactions Involving Gases The gas laws can be applied to calculate the stoichiometry of reactions in which gases are reactants or products. 2H 2 (g) + O 2 (g) → 2H 2 O(g) 2 mol H 2 reacts with 1 mol O 2 to produce 2 mol water vapor.

Stoichiometry and Volume-Volume Problems Coefficients in a balanced equation represent volume ratios for gases.

Stoichiometry and Volume-Mass Problems Mass must be found by converting to moles or volumes. Plastics are some of the products created with polymers. One component of polymers is ethene gas, or ethylene.

A. A B. B C. C D. D Section 13.3 Assessment How many mol of hydrogen gas are required to react with 1.50 mol oxygen gas in the following reaction? 2H 2 (g) + O 2 (g) → 2H 2 O(g) A.1.00 B.2.00 C.3.00 D.4.00

A. A B. B C. C D. D Section 13.3 Assessment How many liters of hydrogen gas are required to react with 3.25 liters of oxygen gas in the following reaction? 2H 2 (g) + O 2 (g) → 2H 2 O(g) A.2.00 B.3.25 C.4.00 D.6.50

SAMPLE PROBLEM 1 CH 4 burns in O 2, producing CO 2 and H 2 O(g). A 1.22 L CH 4 cylinder, at 15°C, registers a pressure of 328 kPa. a)What volume of O 2 at SATP 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 (8.31 kPaL/Kmol)(288 K) (328 kPa)(1.22 L) = n = mol P = 328 kPa, V = 1.22 L, T = 288 K # mol O 2 = mol CH 4 2 mol O 2 1 mol CH 4 x = mol PV = nRT (100 kPa) (0.334 mol)(8.31 kPaL/Kmol )(298 K) =V = 8.28 L P= 100 kPa, n= mol, T= 298 K or # L = mol x 24.8 L/mol = 8.28 L

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? c)What volume of CO 2 (at STP) is produced if only 2.15 g of the CH 4 was burned? # g H 2 O= mol CH 4 2 mol H 2 O 1 mol CH 4 x = 6.02 g H 2 O g H 2 O 1 mol H 2 O x # mol CO 2 = 2.15 g CH 4 1 mol CH g CH 4 x = mol CO 2 1 mol CO 2 1 mol CH 4 x PV = nRTP = kPa, n = mol, T = 273 K (101.3 KPa) (0.134 mol)(8.31 kPaL/Kmol )(273 K) = V = 3.00 L CO 2 or # L = 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? # mol NH 3 = 7500 g H 2 1 mol H g H 2 x = 2475 mol 2 mol NH 3 3 mol H 2 x PV = nRTP = 450 kPa, n = 2475 mol, T = 353 K (450 KPa) (2475 mol)(8.31)(353 K) = V = L NH 3 First we need a balanced equation: N 2 (g) + 3H 2 (g)  2NH 3 (g)

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) # g Na= mol H 2 = 41.1 g Na 2 mol Na 1 mol H 2 x g Na 1 mol Na x PV = nRT (8.31 kPaL/Kmol )(273 K) (101.3 kPa)(20.0 L) = n = mol H 2 P= kPa, V= 20.0 L, T= 273 K or # mol = 20.0 L x 1 mol / 22.4 L = mol

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 OH 1 mol CH 3 OH g CH 3 OH x = 15.7 L O 2 3 mol O 2 2 mol CH 3 OH x 22.4 L O 2 1 mol O 2 x 2. 2NaCl(l)  2Na(l) + Cl 2 (g) # mol Cl 2 =105 g Na 1 mol Na g Na x 1 mol Cl 2 2 mol Na x PV = nRTP = 100 kPa, n = mol, T = 1073 K (100 KPa) (2.284 mol)(8.31)(1073 K) = V = 204 L Cl 2 = mol 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 1 mol O L O 2 x = 7.1 g C 3 H 8 1 mol C 3 H 8 5 mol O 2 x g C 3 H 8 1 mol C 3 H 8 x 4. C 3 H 8 (g) + 5O 2 (g)  3CO 2 (g) + 4H 2 O(g) PV = nRT # mol O 2 =2.8 mol C 3 H 8 5 mol O 2 1 mol C 3 H 8 x = 14 mol O 2 (8.31)(283 K) (1317 kPa)(5.0 L) n == 2.8 mol C 3 H 8 PV = nRTP = 103 kPa, n = 14 mol, T = 283 K (103 KPa) (14 mol)(8.31)(283 K) = V = 320 L O 2

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

5. # mol C 3 H 5 (NO 3 ) 3 = 200 g C 3 H 5 (NO 3 ) 3 1 mol C 3 H 5 (NO 3 ) g C 3 H 5 (NO 3 ) 3 x = mol # mol all gases= mol C 3 H 5 (NO 3 ) 3 29 mol gases 4 mol C 3 H 5 (NO 3 ) 3 x = mol all gases PV = nRTV = 50 L, n = mol, T = 493 K (50 L) (6.385 mol)(8.31)(493 K) = P = 523 kPa For more lessons, visit