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BellaGrey
@BellaGrey
August 2022
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Какая масса аммиака образуется при взаимодействии 600 л водорода с азотом
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DarkChemist
N2 + 3H2 -> 2NH3
n(H2) = V / Vm = 600 / 22,4 = 26,8 (моль)
n(NH3) = n(H2) / 3 * 2 = 17,9 (моль)
m(NH3) = n * M = 17,9 * 17 = 304 (г)
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Answers & Comments
n(H2) = V / Vm = 600 / 22,4 = 26,8 (моль)
n(NH3) = n(H2) / 3 * 2 = 17,9 (моль)
m(NH3) = n * M = 17,9 * 17 = 304 (г)