m(Mg) = 24 g
m(MgO практического) = 35 g
η(MgO) - ?
n(Mg) = 24 g / 24 g / mol = 1 mol
2Mg + O2 = 2MgO
2n(Mg) = 2n(MgO) = 1 mol;
m(MgO теоретического) = 1 mol * 40 g / mol = 40 g ;
η(MgO) = 35 g / 40 g = 0.875 = 87.5% ;
Ответ: 87.5% .
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Answers & Comments
m(Mg) = 24 g
m(MgO практического) = 35 g
η(MgO) - ?
n(Mg) = 24 g / 24 g / mol = 1 mol
2Mg + O2 = 2MgO
2n(Mg) = 2n(MgO) = 1 mol;
m(MgO теоретического) = 1 mol * 40 g / mol = 40 g ;
η(MgO) = 35 g / 40 g = 0.875 = 87.5% ;
Ответ: 87.5% .