Предмет: Химия, автор: lera50614

помогите решить. У меня не получается.

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Автор ответа: zubareva45

1

Пусть масса раствора серной кислоты будет 100 г.

Тогда масса воды 22 г.

n(H2O) = 22÷18 =11÷9 моль

H₂O + SO₃ = H₂SO₄

n(SO₃) = n(H₂O) = 11÷9 моль

m(SO₃) = (11÷9)×80 = 880÷9 г

m(олеума) = (880÷9)×10 = 8800÷9

m(раствора серной кислоты)÷m(олеума) = 900÷8800 = 9÷88

Ответ: 9 : 88

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