Ta có: \(m_{ddH_2SO_4}=60.1,8=108\left(g\right)\)
`=>` \(n_{H_2SO_4}=\dfrac{108.85\%}{98}=\dfrac{459}{490}\left(mol\right)\)
`=>` \(n_{H_2SO_4\left(pư\right)}=\dfrac{459}{490}.42,7\%=0,4\left(mol\right)\)
PTHH:
\(CO+CuO\xrightarrow[]{t^o}Cu+CO_2\) (1)
\(Cu+2H_2SO_{4\left(đ\right)}\xrightarrow[]{t^o}CuSO_4+SO_2+2H_2O\) (2)
\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\) (3)
Ta có: \(n_{CuO\left(bđ\right)}=\dfrac{20}{80}=0,25\left(mol\right)\)
Gọi \(n_{CuO\left(pư\right)}=a\left(mol\right)\left(a\le0,25\right)\)
Theo PT (1): \(n_{CO}=n_{Cu}=n_{CuO\left(pư\right)}=a\left(mol\right)\)
`=>` \(n_{CuO\left(dư\right)}=0,25-a\left(mol\right)\)
Theo PT (2), (3): \(n_{H_2SO_4}=2n_{Cu}+n_{CuO\left(dư\right)}\)
`=> 2a + 0,25 - a = 0,4`
`=> a = 0,15 (t//m)`
`=>` \(\left\{{}\begin{matrix}\%V_{CO}=\dfrac{0,15.22,4}{5,6}.100\%=60\%\\\%V_{CO_2}=100\%-60\%=40\%\end{matrix}\right.\)
