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Gas Stoichiometry

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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

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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

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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

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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)

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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

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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.

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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?

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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

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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

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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

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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

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