Acrylonitrile, C3H3N, is the starting material for the production of a kind of synthetic fiber (acrylics) and can be made from propylene, C3H6, when it reacts with nitric oxide, NO, as follows:
4C3H6(g) + 6NO(g) → 4C3H3N(s) + 6H2O(l) + N2(g)
What mass of C3H3N can be made when 25.6 g of C3H6 reacts with 25.6 g of nitric oxide?
chemistry..?
2016-06-13 3:29 pm
回答 (1)
2016-06-13 4:41 pm
Molar mass of C₃H₆ = (12.01×3 + 1.008×6) g/mol = 42.08 g/mol
Molar mass of NO = (14.01 + 16.00) g/mol = 30.01 g/mol
Molar mass of C₃H₃N = (12.01×3 + 1.008×3 + 14.01) g/mol = 53.06 g/mol
4C₃H₆(g) + 6NO(g) → 4C₃H₃N(s) + 6H₂O(l) + N₂(g)
If 25.6 g of C₃H₆ completely reacts :
No. of moles of C₃H₆ reacted = (25.6 g) / (42.08 g/mol) = 25.6/42.08 g/mol
No. of moles of C₃H₃N formed = (25.6/42.08 g/mol) × (4/4)
Mass of C₃H₃N formed = (25.6/42.08 g/mol) × (4/4) × (53.06 g/mol) = 32.3 g
If 25.6 g of NO completely reacts :
No. of moles of C₃H₆ reacted = (25.6 g) / (30.01 g/mol) = 25.6/30.01 g/mol
No. of moles of C₃H₃N formed = (25.6/30.01 g/mol) × (4/6)
Mass of C₃H₃N formed = (25.6/30.01 g/mol) × (4/6) × (53.06 g/mol) = 30.2 g < 32.3 g
Hence, the limiting reactant is NO.
Mass of C₃H₃N formed = 30.2 g
Molar mass of NO = (14.01 + 16.00) g/mol = 30.01 g/mol
Molar mass of C₃H₃N = (12.01×3 + 1.008×3 + 14.01) g/mol = 53.06 g/mol
4C₃H₆(g) + 6NO(g) → 4C₃H₃N(s) + 6H₂O(l) + N₂(g)
If 25.6 g of C₃H₆ completely reacts :
No. of moles of C₃H₆ reacted = (25.6 g) / (42.08 g/mol) = 25.6/42.08 g/mol
No. of moles of C₃H₃N formed = (25.6/42.08 g/mol) × (4/4)
Mass of C₃H₃N formed = (25.6/42.08 g/mol) × (4/4) × (53.06 g/mol) = 32.3 g
If 25.6 g of NO completely reacts :
No. of moles of C₃H₆ reacted = (25.6 g) / (30.01 g/mol) = 25.6/30.01 g/mol
No. of moles of C₃H₃N formed = (25.6/30.01 g/mol) × (4/6)
Mass of C₃H₃N formed = (25.6/30.01 g/mol) × (4/6) × (53.06 g/mol) = 30.2 g < 32.3 g
Hence, the limiting reactant is NO.
Mass of C₃H₃N formed = 30.2 g
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