Предмет: Химия, автор: medzhitovedemp91j3b

Решите задачу на выход

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Автор ответа: UJIJIIO3UOHUCT

Al4C3 + 12H2O → 4Al(OH)3 + 3CH4↑

m(Al4C3) без примесей = 80*(100-8)/100 = 73,6 г

n(Al4C3) = 73,6/144 ≈ 0,5111 моль

По уравнению реакции n(CH4) = 3*n(Al4C3) = 3*0,5111 = 1,5333 моль

С учетом выхода реакции n(CH4) = 1,5333*0,94 = 1,4413 моль

V(CH4) = 1,4413*22,4 ≈ 32,3 л

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