a,
$CuO+CO\xrightarrow{{t^o}} Cu+CO_2$
$CO_2+Ba(OH)_2\to BaCO_3+H_2O$
b,
$n_{CO}=\dfrac{V}{22,4}(mol)$
$n_{CuO}=\dfrac{m}{80}(mol)$
$n_{BaCO_3}=\dfrac{a}{197}(mol)$
$\to n_{CO_2}=n_{BaCO_3}=\dfrac{a}{197}(mol)$
$\to n_{Cu}=n_{CO_2}=\dfrac{a}{197}(mol)$
- Nếu $\dfrac{V}{22,4}\le \dfrac{m}{80}\to 80V\le 22,4m$:
Theo lí thuyết, $CO$ hết, tạo $\dfrac{V}{22,4}(mol)$ $Cu$
$\to H=\dfrac{ \dfrac{a}{197}.100}{\dfrac{V}{22,4}}=11,37.\dfrac{a}{V}\%$
- Nếu $\dfrac{V}{22,4}>\dfrac{m}{80}\to 80V>22,4m$:
Theo lí thuyết, $CuO$ hết, tạo $\dfrac{m}{80}(mol)$ $Cu$
$\to H=\dfrac{\dfrac{a}{197}.100}{\dfrac{m}{80}}=40,6.\dfrac{a}{m}\%$