a) CuO + H2SO4 → CuSO4 + H2O
b) \(n_{CuO}=\frac{16}{80}=0,2\left(mol\right)\)
\(m_{H_2SO_4}=147\times20\%=29,4\left(g\right)\)
\(n_{H_2SO_4}=\frac{29,4}{98}=0,3\left(mol\right)\)
Theo pT: \(n_{CuO}=n_{H_2SO_4}\)
Theo bài: \(n_{CuO}=\frac{2}{3}n_{H_2SO_4}\)
Vì \(\frac{2}{3}< 1\) ⇒ H2SO4 dư
Dung dịch sau pư gồm: H2SO4 dư và CuSO4
\(m_{dd}saupư=16+147=163\left(g\right)\)
Theo PT: \(n_{H_2SO_4}pư=n_{CuO}=0,2\left(mol\right)\)
\(\Rightarrow n_{H_2SO_4}dư=0,3-0,2=0,1\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}dư=0,1\times98=9,8\left(g\right)\)
\(\Rightarrow C\%_{H_2SO_4}dư=\frac{9,8}{163}\times100\%=6,01\%\)
Theo PT: \(n_{CuSO_4}=n_{CuO}=0,2\left(mol\right)\)
\(\Rightarrow m_{CuSO_4}=0,2\times160=32\left(g\right)\)
\(\Rightarrow C\%_{CuSO_4}=\frac{32}{163}\times100\%=19,63\%\)
