Đáp án:

a)

$2Al+6HCl\to2AlCl_3+3H_2$

$Fe+2HCl→FeCl_2+H_2$

b)

$m_{Al} = 2,7 gam,\,  m_{Fe} = 5,6 gam$

c) V=250 ml

d) 0,4 và 0,4

Giải thích các bước giải

$\begin{array}{l}
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
x\,\,\,\,\,\,\,\, \to 3x \to \,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{3}{2}x\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
y\,\,\,\,\,\,\,\,\,\,\,\,\,\,2y \to \,\,\,\,\,y \to \,\,\,\,\,\,\,\,y
\end{array}$

$\left\{ \begin{array}{l}
27x + 56y = 8,3\\
\frac{3}{2}x + y = 0,25
\end{array} \right. \to \left\{ \begin{array}{l}
x = 0,1\\
y = 0,1
\end{array} \right.$

b)

$m_{Al}= 0,1.27 = 2,7gam$

$m_{Fe} = 0,1.56 = 5,6gam$

c)

$n_{HCl} = 3.0,1 + 2.0,1 = 0,5\,\,mo$

$V_{HCl}= \frac{{0,5}}{2} = 0,25\,\,l = 250\,\,ml$

d)

$C_{M\,AlC{l_3}} = \frac{{0,1}}{{0,25}} = 0,4M$

$C_{M\,AlC{l_3}} = \frac{{0,1}}{{0,25}} = 0,4M$