21.
Beginning with 44.0 g of CaH2 and 46.0 g of H2O, what volume of H2 will be produced at standard temperature and pressure?
Calcium hydride combines with water, according to the following equation:
CaH2(s) + 2H2O(l) → 2H2(g) + Ca(OH)2(s)
57.2 L
46.1 L
23.0 L
103 L
22.
Nitroglycerin is an explosive used by the mining industry. It detonates according to the following balanced equation:
4C3H5N3O9(l) → 12CO2(g) + 6N2(g) + 10H2O(g) + O2(g)
What volume is occupied by the carbon dioxide gas produced when 12.0 g of nitroglycerine explodes? The pressure is 1.25 atm at 523°C
0.370 L
8.29 L
2.76 L
1.52 L
23.
The burning of ethane produces both CO2 and H2O. If 600 mL of CO2 is produced at 25°C and 730 torr, what volume of water vapor would be produced at 34°C and 810 torr? Use the equation:
C2H6 + O2 → CO2 + H2O
(The equation needs to be balanced.)
0.836 L
1.03 L
0.525 L
0.787 L
24.
The burning of ethane produces both CO2 and H2O. If 400 mL of CO2 is produced at 30°C and 740 torr, what volume of water vapor would be produced at 19°C and 780 torr? Use the equation:
C2H6 + O2 → CO2 + H2O
(The equation needs to be balanced.)
0.591 L
0.656 L
0.366 L
0.549 L
25.
If 10.0 grams of KOH reacts with 25.0 grams (NH4)2SO4 according to the following reaction, calculate the liters of NH3 produced at 500 torr and 20°C.
2KOH + (NH4)2SO4 → K2SO4 + 2NH3 + 2H2O
(The equation is already balanced.)
13.8 L
6.51 L
2.45 L
20.3 L
5 chem review questions?
2018-01-13 3:24 pm
回答 (1)
2018-01-13 5:30 pm
21.
Initial no. of moles of CaH₂ added = (44.0 g) / [(40.1 + 1.0×2) g/mol] = 1.045 mol
Initial no. of moles of H₂O added = (46.0 g) / [(1.0×2 + 16) g/mol] = 2.56 mol
CaH₂(s) + 2H₂O(l) → 2H₂(g) + Ca(OH)₂(s)
Mole ratio CaH₂ : H₂O : H₂ = 1 : 2 : 2
When 1.05 mol CaH₂ completely reacts, H₂O needed = (1.045 mol) × (2/1) = 2.09 mol < 2.56 mol
H₂O is in excess, and thus CaH₂ is the limiting reactant/reagent which completely reacts.
No. of moles of CaH₂ reacted = 1.045 mol
No. of moles of H₂ gas produced at STP = (1.045 mol) × 2 = 2.09 mol
Each moles of gas occupies a volume of 22.4 L at STP.
Volume of H₂ produced at STP = (2.09 mol) × (22.4 L/mol) = 46.8 L ≈ 46.1 L
The answer : 46.1 L
====
22.
Molar mass of C₃H₅N₃O₉ = (12.0×3 + 1.0×5 + 14.0×3 + 16.0×9) g/mol = 227.0 g/mol
No. of moles of C₃H₅N₃O₉ = (12.0 g) / [(12.0×3 + 1.0×5 + 14.0×3 + 16.0×9) g/mol] = 0.05286 mol
4C₃H₅N₃O₉(l) → 12CO₂(g) + 6N₂(g) + 10H₂O(g) + O₂(g)
Mole ratio C₃H₅N₃O₉ : CO₂ = 4 : 12
No. of moles of CO₂ produced = (0.05286 mol) × (12/4) = 0.1586 mol
Gas law : PV = nRT
Volume of CO₂, V = nRT/P = 0.1586 × 0.08206 × (273 + 523) / 1.25 L = 8.29 L
The answer : 8.29 L
====
23.
Gas law : PV= nRT
No. of moles of CO₂ produced, n = PV/(RT) = (730/760) × (600/1000) / [0.08206 × (273 + 25)] mol = 0.02357 mol
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
Mole ratio CO₂ : H₂O = 4 : 6
No. of moles of H₂O vapor produced = (0.02357 mol) × (6/4) = 0.03536 mol
Gas law : PV = nRT
Volume of H₂O vapor, V = nRT/P = 0.03536 × 0.08206 × (273 + 34) / (810/760) L = 0.836 L
The answer : 0.836 L
====
24.
Gas law : PV= nRT
No. of moles of CO₂ produced, n = PV/(RT) = (740/760) × (400/1000) / [0.08206 × (273 + 30)] mol = 0.01566 mol
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
Mole ratio CO₂ : H₂O = 4 : 6
No. of moles of H₂O vapor produced = (0.01566 mol) × (6/4) = 0.02349 mol
Gas law : PV = nRT
Volume of H₂O vapor, V = nRT/P = 0.02349 × 0.08206 × (273 + 19) / (780/760) L = 0.548 L ≈ 0.549 L
The answer : 0.549 L
====
25.
Initial no. of moles of KOH added = (10.0 g) / [(39.0 + 16.0 + 1.0) g/mol] = 0.1786 mol
Initial no. of moles of (NH₄)₂SO₄ added = (25.0 g) / [(14.0×2 + 1.0×8 + 32.1 + 16.0×4) g/mol] = 0.1893 mol
2KOH + (NH₄)₂SO₄ → K₂SO₄ + 2NH₃ + 2H₂O
Mole ratio KOH : (NH₄)₂SO₄ : NH₃ = 2 : 1 : 2
When 0.1786 mol KOH complete reacts, (NH₄)₂SO₄ needed = (0.1786 mol) × (1/2) = 0.0893 mol < 0.1893 mol
(NH₄)₂SO₄ is in excess, and thus KOH completely reacts.
No. of moles of KOH reacted = 0.1785 mol
No. of moles of NH₃ produced = (0.1785 mol) × (2/2) = 0.1785 mol
Gas law : PV = nRT
Volume of NH₃ gas, V = nRT/P = 0.1785 × 0.08206 × (273 + 20) / (500/760) L = 6.52 L ≈ 0.651 L
The answer : 0.651 L
Initial no. of moles of CaH₂ added = (44.0 g) / [(40.1 + 1.0×2) g/mol] = 1.045 mol
Initial no. of moles of H₂O added = (46.0 g) / [(1.0×2 + 16) g/mol] = 2.56 mol
CaH₂(s) + 2H₂O(l) → 2H₂(g) + Ca(OH)₂(s)
Mole ratio CaH₂ : H₂O : H₂ = 1 : 2 : 2
When 1.05 mol CaH₂ completely reacts, H₂O needed = (1.045 mol) × (2/1) = 2.09 mol < 2.56 mol
H₂O is in excess, and thus CaH₂ is the limiting reactant/reagent which completely reacts.
No. of moles of CaH₂ reacted = 1.045 mol
No. of moles of H₂ gas produced at STP = (1.045 mol) × 2 = 2.09 mol
Each moles of gas occupies a volume of 22.4 L at STP.
Volume of H₂ produced at STP = (2.09 mol) × (22.4 L/mol) = 46.8 L ≈ 46.1 L
The answer : 46.1 L
====
22.
Molar mass of C₃H₅N₃O₉ = (12.0×3 + 1.0×5 + 14.0×3 + 16.0×9) g/mol = 227.0 g/mol
No. of moles of C₃H₅N₃O₉ = (12.0 g) / [(12.0×3 + 1.0×5 + 14.0×3 + 16.0×9) g/mol] = 0.05286 mol
4C₃H₅N₃O₉(l) → 12CO₂(g) + 6N₂(g) + 10H₂O(g) + O₂(g)
Mole ratio C₃H₅N₃O₉ : CO₂ = 4 : 12
No. of moles of CO₂ produced = (0.05286 mol) × (12/4) = 0.1586 mol
Gas law : PV = nRT
Volume of CO₂, V = nRT/P = 0.1586 × 0.08206 × (273 + 523) / 1.25 L = 8.29 L
The answer : 8.29 L
====
23.
Gas law : PV= nRT
No. of moles of CO₂ produced, n = PV/(RT) = (730/760) × (600/1000) / [0.08206 × (273 + 25)] mol = 0.02357 mol
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
Mole ratio CO₂ : H₂O = 4 : 6
No. of moles of H₂O vapor produced = (0.02357 mol) × (6/4) = 0.03536 mol
Gas law : PV = nRT
Volume of H₂O vapor, V = nRT/P = 0.03536 × 0.08206 × (273 + 34) / (810/760) L = 0.836 L
The answer : 0.836 L
====
24.
Gas law : PV= nRT
No. of moles of CO₂ produced, n = PV/(RT) = (740/760) × (400/1000) / [0.08206 × (273 + 30)] mol = 0.01566 mol
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
Mole ratio CO₂ : H₂O = 4 : 6
No. of moles of H₂O vapor produced = (0.01566 mol) × (6/4) = 0.02349 mol
Gas law : PV = nRT
Volume of H₂O vapor, V = nRT/P = 0.02349 × 0.08206 × (273 + 19) / (780/760) L = 0.548 L ≈ 0.549 L
The answer : 0.549 L
====
25.
Initial no. of moles of KOH added = (10.0 g) / [(39.0 + 16.0 + 1.0) g/mol] = 0.1786 mol
Initial no. of moles of (NH₄)₂SO₄ added = (25.0 g) / [(14.0×2 + 1.0×8 + 32.1 + 16.0×4) g/mol] = 0.1893 mol
2KOH + (NH₄)₂SO₄ → K₂SO₄ + 2NH₃ + 2H₂O
Mole ratio KOH : (NH₄)₂SO₄ : NH₃ = 2 : 1 : 2
When 0.1786 mol KOH complete reacts, (NH₄)₂SO₄ needed = (0.1786 mol) × (1/2) = 0.0893 mol < 0.1893 mol
(NH₄)₂SO₄ is in excess, and thus KOH completely reacts.
No. of moles of KOH reacted = 0.1785 mol
No. of moles of NH₃ produced = (0.1785 mol) × (2/2) = 0.1785 mol
Gas law : PV = nRT
Volume of NH₃ gas, V = nRT/P = 0.1785 × 0.08206 × (273 + 20) / (500/760) L = 6.52 L ≈ 0.651 L
The answer : 0.651 L
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