CuO + H2SO4 → CuSO4 + H2O
\(n_{CuO}=\dfrac{3,2}{80}=0,04\left(mol\right)\)
Theo PT: \(n_{H_2SO_4}=n_{CuO}=0,04\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}=0,04\times98=3,92\left(g\right)\)
\(\Rightarrow C\%_{ddH_2SO_4}=\dfrac{3,92}{100}\times100\%=3,92\%\)
Vậy X%=3,92%
Theo PT: \(n_{CuSO_4}=n_{CuO}=0,04\left(mol\right)\)
\(\Rightarrow m_{CuSO_4}=0,04\times160=6,4\left(g\right)\)
\(\Sigma m_{dd}=3,2+100=103,2\left(g\right)\)
\(\Rightarrow C\%_{ddCuSO_4}=\dfrac{6,4}{103,2}\times100\%=6,2\%\)