$m_{dd\,giảm}=m_O=16.25\%=4(g)$
$\xrightarrow{\rm Bảo\,toàn\,O: }n_{H_2O}=n_O=\dfrac{4}{16}=0,25(mol)$
Đặt $n_{CuO}=x(mol);n_{Fe_2O_3}=y(mol)$
$\to 80x+160y=16(1)$
$CuO+H_2\xrightarrow{t^o}Cu+H_2O$
$Fe_2O_3+3H_2\xrightarrow{t^o}2Fe+3H_2O$
$\xrightarrow{\rm Theo\,PT: }x+3y=0,25(2)$
Từ $(1)(2)\to \begin{cases} x=0,1\\ y=0,05 \end{cases}$
$\to \%m_{CuO}=\dfrac{0,1.80}{16}.100\%=50\%$
$\to \%m_{Fe_2O_3}=100-50=50\%$
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