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tamara20010206
@tamara20010206
August 2021
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6 грамм смеси Al и Cu обработали серной кислотой, при этом выделилось 6.72 л H2.
Определить процентный состав Cu
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Potassium
Cu + H2SO4 ≠
2Al + 3H2SO4 = Al2(SO4)3 + 3H2
x/54 = 6.72/67.2
x = 5.4 грамма Al.
==> m(Cu) = 6 - 5.4 = 0.6 грамм ==>
W(Cu) = (0.6/6)*100% = 10%>
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Answers & Comments
2Al + 3H2SO4 = Al2(SO4)3 + 3H2
x/54 = 6.72/67.2
x = 5.4 грамма Al.
==> m(Cu) = 6 - 5.4 = 0.6 грамм ==>
W(Cu) = (0.6/6)*100% = 10%>