Предмет: Химия, автор: ригина

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№1

 Какую массу соли можно получить в реакции магния (избыток) из серной кислоты массой 24,5 г

Ответы

Автор ответа: AnnKalevich
0

        0,25 моль  х моль

Mg + H2SO4 = MgSO4 + H2

       1 моль    1 моль

n = m/M

M(H2SO4) = 1*2 + 32 + 16*4 = 98 г/моль

n(H2SO4) = 24,5 г / 98 г/моль = 0,25 моль

0,25 = х

   1      1

х = 0,25

n(MgSO4) = 0,25 моль

m = n*M

M(MgSO4) = 24 + 32 + 64 = 120 г/моль

m(MgSO4) = 0,25 моль * 120 г/моль = 30 г

Ответ: 30 г

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