Đáp án + Giải thích các bước giải:

$a)PTHH:$

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

$Mg+2HCl\to MgCl_2+H_2\uparrow$

$\Rightarrow m_{Cu}=m_{\text{rắn ko tan}}=9,6(g)$

Đặt $n_{Al}=x(mol);n_{Mg}=y(mol)$

Ta có $m_{Al}+m_{Mg}=m_{hh}-m_{Cu}$

$\Rightarrow 27x+24y=14,7-9,6=5,1(g)$

$n_{H_2}=\dfrac{5,6}{22,4}=0,25(mol)$

Theo PT: $1,5n_{Al}+n_{Mg}=n_{H_2}$

$\Rightarrow 1,5x+y=0,25(2)$

Từ $(1)(2)\Rightarrow x=y=0,1(mol)$

$\Rightarrow \begin{cases} m_{Al}=0,1.27=2,7(g)\\ m_{Mg}=0,1.24=2,4(g)\\ m_{Cu}=9,6(g)\end{cases}$

$b)$ Theo PT: $\Sigma n_{HCl}=2\Sigma n_{H_2}=0,5(mol)$

$\Rightarrow n_{HCl(dùng)}=0,5.(100\%+20\%)=0,6(mol)$

Đổi $200ml=0,2l$

$\Rightarrow C_{M_{HCl}}=\dfrac{0,6}{0,2}=3M$

$c)$ Theo PT: $n_{AlCl_3}=n_{MgCl_2}=0,1(mol)$

$\Rightarrow C_{M_{AlCl_3}}=C_{M_{MgCl_2}}=\dfrac{0,1}{0,2}=0,5M$

$d)PTHH:Cu+Cl_2\xrightarrow{t^o}CuCl_2$

$n_{Cu}=\dfrac{9,6}{64}=0,15(mol)$

Theo PT: $n_{Cl_2}=n_{Cu}=0,15(mol)$

$\Rightarrow V_{Cl_2(đktc)}=0,15.22,4=3,36(l)$

Đáp án:

$a,\\ m_{Al}=2,7(g)\\ m_{Mg}=2,4(g)\\ m_{Cu}=9,6(g)\\ b,\ CM_{dd\ HCl}=3(M)\\ c,\ CM_{dd AlCl_3}=CM_{dd\ MgCl_2}=0,5(M)\\ d,\ V_{Cl_2}=3,36(l)$

Giải thích các bước giải:

$a,\\ m_{Cu}=9,6(g)\\ n_{H_2}=\dfrac{5,6}{22,4}=0,25(mol)\\ Đặt\ \begin{cases}n_{Al}=x(mol)\\n_{Mg}=y(mol)\end{cases}\\ \Rightarrow 27x+24y=14,7-9,6=5,1\ (1)\\ 2Al+6HCl\to 2AlCl_3+3H_2\\ Mg+2HCl\to MgCl_2+H_2\\ \xrightarrow{\text{Từ PTHH}}\ n_{H_2}=\dfrac{3}{2}.n_{Al}+n_{Mg}=\dfrac{3}{2}.x+y(mol)\\ \Rightarrow \dfrac{3}{2}.x+y=0,25\ (2)\\ \xrightarrow{\text{Từ (1);(2)}}\ \begin{cases}x=0,1\\y=0,1\end{cases}\\ \Rightarrow \begin{cases}m_{Al}=0,1.27=2,7(g)\\m_{Mg}=0,1.24=2,4(g)\\m_{Cu}=9,6(g)\end{cases}\\ b,\\ \xrightarrow{\text{Từ PTHH}}\ n_{HCl\ \text{phản ứng}}=2.n_{H_2}=0,5(mol)\\ n_{HCl\ \text{thực tế}}=0,5.120\%=0,6(mol)\\ CM_{dd\ HCl}=\dfrac{0,6}{0,2}=3(M)\\ c,\\ \xrightarrow{\text{Từ PTHH}}\ \begin{cases}n_{AlCl_3}=n_{Al}=0,1(mol)\\n_{MgCl_2}=n_{Mg}=0,1(mol)\end{cases}\\ \Rightarrow CM_{dd\ AlCl_3}=CM_{dd\ MgCl_2}=\dfrac{0,1}{0,2}=0,5(M)\\ d,\\ n_{Cu}=\dfrac{9,6}{64}=0,15(mol)\\ Cu+Cl_2\xrightarrow{t^0} CuCl_2\\ \xrightarrow{\text{Từ PTHH}}\ n_{Cl_2}=n_{Cu}=0,15(mol)\\ V_{Cl_2}=0,15.22,4=3,36(l)$