Giải thích các bước giải:

$a)$

Đặt: $n_{Zn}=x(mol),\ n_{Al}=y(mol)$

$65x+27y=17,05(g)\ (1)$

$Zn+2HCl\to ZnCl_2+H_2$

$2Al+6HCl\to 2AlCl_3+3H_2$

$n_{H_2}=\dfrac{9,52}{22,4}=0,425(mol)$

$x+1,5y=0,425(mol)\ (2)$

Từ $(1)(2)$ giải được:

$x=0,2(mol),\ y=0,15(mol)$

$m_{Zn}=0,2×65=13(g)$

$m_{Al}=0,15×27=4,05(g)$

$b)$

Theo phương trình:

$n_{ZnCl_2}=n_{Zn}=0,2(mol)$

$n_{AlCl_3}=n_{Al}=0,15(mol)$

$n_{HCl\ \rm (pứ)}=2n_{H_2}=0,85(mol)$

Đổi: $500ml=0,5l$

$n_{HCl}=0,5×2=1(mol)$

$n_{HCl\ \rm (dư)}=1-0,85=0,15(mol)$

$C_{M\ HCl\ \rm (dư)}=\dfrac{0,15}{0,5}=0,3M$

$C_{M\ ZnCl_2}=\dfrac{0,2}{0,5}=0,4M$

$C_{M\ AlCl_3}=\dfrac{0,15}{0,5}=0,3M$

$c)$

$m_{dd\ HCl}=500×1,05=525(g)$

$m_{dd\ sau}=17,05+525-0,85=541,2(g)$

$C\%_{HCl\ \rm (dư)}=\dfrac{0,15×36,5}{541,2}×100\%=1,01\%$

$C\%_{ZnCl_2}=\dfrac{0,2×136}{541,2}×100\%=5,026\%$

$C\%_{AlCl_3}=\dfrac{0,15×133,5}{541,2}×100\%=3,7\%$

Đáp án:

$a\big)\begin{cases} m_{Zn}=13(g)\\ m_{Al}=4,05(g) \end{cases}$

$b\big)\begin{cases} C_{M\,HCl(dư)}=0,3M\\ C_{M\,ZnCl_2}=0,4M\\ C_{M\,AlCl_3}=0,3M \end{cases}$

$c\big)\begin{cases} C\%_{HCl(dư)}≈1,01\%\\ C\%_{ZnCl_2}≈5,03\%\\ C\%_{AlCl_3}≈3,7\% \end{cases}$

Giải thích các bước giải:

$a\big)$

Đặt $n_{Zn}=x(mol);n_{Al}=y(mol)$

$\to 65x+27y=17,05(1)$

Đổi $500ml=0,5l$

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

$n_{H_2}=\frac{9,52}{22,4}=0,425(mol)$

$Zn+2HCl\to ZnCl_2+H_2$

$2Al+6HCl\to 2AlCl_3+3H_2$

Theo PT: $x+1,5y=n_{H_2}=0,425(2)$

Từ $(1)(2)\to\begin{cases} x=0,2\\ y=0,15 \end{cases}$

$\to\begin{cases} m_{Zn}=0,2.65=13(g)\\ m_{Al}=0,15.27=4,05(g) \end{cases}$

$b\big)$

Theo PT: $n_{HCl(pứ)}=2n_{H_2}=0,85(mol)$

$\to n_{HCl(dư)}=1-0,85=0,15(mol)$

Theo PT: $n_{ZnCl_2}=n_{Zn}=0,2(mol);n_{AlCl_3}=n_{Al}=0,15(mol)$

$\to \begin{cases} C_{M\,HCl(dư)}=\frac{0,15}{0,5}=0,3M\\ C_{M\,ZnCl_2}=\frac{0,2}{0,5}=0,4M\\ C_{M\,AlCl_3}=\frac{0,15}{0,5}=0,3M \end{cases}$

$c\big)$

$m_{dd\,HCl}=1,05.500=525(g)$

Ta có $m_{dd\,spu}=17,05+525-0,425.2=541,2(g)$

$\to \begin{cases} C\%_{HCl(dư)}=\frac{0,15.36,5}{541,2}.100\%≈1,01\%\\ C\%_{ZnCl_2}=\frac{0,2.136}{541,2}.100\%≈5,03\%\\ C\%_{AlCl_3}=\frac{0,15.133,5}{541,2}.100\%≈3,7\% \end{cases}$