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niginsha2003
@niginsha2003
August 2022
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сколько железа и серной кислоты необходимо для получения 4,48 л водорода
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protector
Fe + H2SO4 = FeSO4 + H2
n(H2) = 4,48/22,4 = 0,2 моль
m (H2SO4) = 0,2*98 = 19,6 г
m(Fe) =0,2*56 = 11,2 г
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niginsha2003
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Answers & Comments
n(H2) = 4,48/22,4 = 0,2 моль
m (H2SO4) = 0,2*98 = 19,6 г
m(Fe) =0,2*56 = 11,2 г