Đáp án + Giải thích các bước giải:

$a\bigg)$

$n_{H_2}=\frac{8,96}{22,4}=0,4(mol)$

$\sum n_{HCl}=0,5.2=1(mol)$

$Zn+2HCl\to ZnCl_2+H_2(1)$

$ZnO+2HCl\to ZnCl_2+H_2O(2)$

Theo PT(1): $n_{Zn}=0,4(mol);n_{HCl(1)}=0,8(mol)$

$\to n_{HCl(2)}=1-0,8=0,2(mol)$

Theo PT(1): $n_{ZnO}=\frac{1}{2}n_{HCl(2)}=0,1(mol)$

$\to m_{Zn}=0,4.65=26(g);m_{ZnO}=0,1.81=8,1(g)$

$\to \%m_{Zn}=\frac{26}{26+8,1}.100\%\approx 76,25\%$

$\to \%m_{ZnO}=100-76,25=23,75\%$

$b\bigg)$

Đổi $500ml=0,5l$

Theo PT: $\sum n_{ZnCl_2}=n_{Zn}+n_{ZnO}=0,5(mol)$

$\to C_{M_{ZnCl_2}}=\frac{0,5}{0,5}=1M$

$c\bigg)$

Đổi $250ml=0,25l$

$n_{HCl}=0,25.2=0,5(mol)$

$n_{Na_2CO_3}=\frac{21,2}{106}=0,2(mol)$

$Na_2CO_3+2HCl\to 2NaCl+CO_2+H_2O$

Xét tỉ lệ: $\frac{n_{Na_2CO_3}}{1}<\frac{n_{HCl}}{2}\to HCl$ dư

Theo PT: $n_{CO_2}=n_{Na_2CO_3}=0,2(mol)$

$\to V_{CO_2}=0,2.22,4=4,48(l)$