Ammonia is produced from the reaction of nitrogen and hydrogen according to the following balanced equation:
N2(g) + 3H2(g) → 2NH3(g)
1. What is the maximum mass of ammonia that can be produced from a mixture of 182.7 g of N2 and 45.74 g of H2?
g
2. Which element would be left partially unreacted? (enter nitrogen or hydrogen)
3. What mass of the starting material would remain unreacted?
g
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2018-03-06 12:15 pm
回答 (1)
2018-03-06 6:34 pm
1.
Molar mass of N₂ = 14.01 × 2 g/mol = 28.02 g/mol
Molar mass of H₂ = 1.008 × 2 g/mol = 2.016 g/mol
Molar mass of NH₃ = (14.01 + 1.008×3) g/mol = 17.03 g/mol
Initial number of moles of N₂ = (182.7 g) / (28.02 g/mol) = 6.520 mol
Initial number of moles of H₂ = (45.74 g) / (2.016 g/mol) = 22.69 mol
Balanced equation for the reaction :
N₂(g) + 3H₂(g) → 2NH₃(g)
Mole ratio N₂ : H₂ : NH₃ = 1 : 3 : 2
If 6.520 mol N₂ completely reacts, H₂ needed = (6.525 mol) × 3 = 19.56 mol < 22.69 mol
H₂ is in excess, and thus the limiting reactant/reagent is N₂.
Number of moles of N₂ reacted = 6.520 mol
Maximum no. of moles of NH₃ produced = (6.520 mol) × 2 = 13.04 mol
Maximum mass NH₃ produced = (13.04 mol) × (17.03 g/mol) = 222.1 g
2.
hydrogen
3.
No. of moles of H₂ left unreacted = (22.69 - 19.56) mol = 3.13 mol
Mass of H₂ left unreacted = (3.13 mol) × (2.016 g/mol) = 6.31 g
Molar mass of N₂ = 14.01 × 2 g/mol = 28.02 g/mol
Molar mass of H₂ = 1.008 × 2 g/mol = 2.016 g/mol
Molar mass of NH₃ = (14.01 + 1.008×3) g/mol = 17.03 g/mol
Initial number of moles of N₂ = (182.7 g) / (28.02 g/mol) = 6.520 mol
Initial number of moles of H₂ = (45.74 g) / (2.016 g/mol) = 22.69 mol
Balanced equation for the reaction :
N₂(g) + 3H₂(g) → 2NH₃(g)
Mole ratio N₂ : H₂ : NH₃ = 1 : 3 : 2
If 6.520 mol N₂ completely reacts, H₂ needed = (6.525 mol) × 3 = 19.56 mol < 22.69 mol
H₂ is in excess, and thus the limiting reactant/reagent is N₂.
Number of moles of N₂ reacted = 6.520 mol
Maximum no. of moles of NH₃ produced = (6.520 mol) × 2 = 13.04 mol
Maximum mass NH₃ produced = (13.04 mol) × (17.03 g/mol) = 222.1 g
2.
hydrogen
3.
No. of moles of H₂ left unreacted = (22.69 - 19.56) mol = 3.13 mol
Mass of H₂ left unreacted = (3.13 mol) × (2.016 g/mol) = 6.31 g
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