3.Cu + 2H2SO4 → CuSO4 + SO2↑ + 2H2O
CuO + H2SO4 → CuSO4 + 2H2O
\(n_{SO_2}=n_{Cu}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
=>\(n_{CuO}=\dfrac{28-0,25.64}{80}=0,15\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{112.70\%}{98}=0,8\left(mol\right)\)
=> \(n_{H_2SO_4\left(dư\right)}=0,8-\left(0,25.2+0,15\right)=0,15\left(mol\right)\)
\(m_{H_2SO_4\left(dư\right)}=0,15.98=14,7\left(g\right)\)
\(m_{CuSO_4}=\left(0,25+0,15\right).160=40\left(g\right)\)
c) \(\%m_{Cu}=\dfrac{0,25.64}{28}=57,14\%\)
=> %mCuO=100 - 57,14 = 42,86%
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