Gọi số mol CuO là x; Mg là y
\(\Rightarrow80x+24y=9,2\)
Ta có:
\(n_{HBr}=0,6.0,5=0,3\left(mol\right)\)
\(CuO+2HBr\rightarrow CuBr_2+H_2O\)
\(Mg+2HBr\rightarrow MgBr_2+H_2\)
\(n_{HBr}=2n_{CuO}+2n_{Mg}=2x+2y=0,3\)
\(\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,05\end{matrix}\right.\)
\(\Rightarrow m_{CuO}=80x=8\left(g\right)\Rightarrow m_{Mg}=1,2\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{CuO}=\frac{8}{9,2}.100\%=86,96\%\\\%m_{Mg}=100\%-86,96\%=13,04\%\end{matrix}\right.\)
\(n_{H2}=n_{Mg}=0,05\left(mol\right)\)
\(\Rightarrow V_{H2}=0,05.22,4=1,12\left(l\right)\)
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