: Cho 10,5 gam hỗn hợp hai kim loại Zn, Cu vào dung dịch H2SO4 loãng dư, người ta thu được 3,36 lít khí (đktc). Khối lượng của kim loại đồng trong hỗn hợp ban đầu là: (Zn = 65, Cu = 64)

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Đáp án:

 $\text{mCu: 0,75g}$

Giải thích các bước giải:

$\text{Ta chỉ có 1 PTHH xảy ra:}$

$\rightarrow$ $\text{Zn+H2SO4$\rightarrow$ZnSO4+H2}$

$\text{nH2=$\frac{V}{22,4}$=$\frac{3,36}{22,4}$=0,15 mol}$

$\text{Có nZn=nH2=0,15 mol}$

$\rightarrow$ $\text{mZn=n.M=0,15.65=9,75g}$

$\text{mCu=10,5-9,75=0,75g}$

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