$Zn+2HCl→ZnCl_2+H_2$
a) $n_{Zn}=\dfrac{6,5}{65}=0,1(mol)$
$n_{H_2}=n_{Zn}=0,1(mol)$
→ $V_{H_2}=0,1×22,4=2,24(l)$
b) $n_{HCl}=2n_{Zn}=2×0,1=0,2(mol)$
→ $m_{HCl}=0,2×36,5=7,3(g)$
c) $CuO+H_2\buildrel{{t^o}}\over\longrightarrow Cu+H_2O$
$n_{CuO}=\dfrac{4}{80}=0,05(mol)$
Xét $n_{H_2}$ và $n_{CuO}$
Ta dễ thấy $0,1>0,05$ ⇒ $CuO$ hết, $H_2$ dư.
Sau phản ứng: $n_{Cu}=n_{H_2}=0,05(mol)$
→ $m_{Cu}=0,05×64=3,2(g)$
