Đáp án:
\(\begin{array}{l}
a)\\
m = 39,4g\\
b)\\
\% {m_{Fe}} = 50,9\% \\
\% {m_{Al}} = 49,1\% \\
c)\\
{C_\% }HCl = 5,84\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
FeO + {H_2} \to Fe + {H_2}O\\
{n_{FeO}} = \dfrac{{28,8}}{{72}} = 0,4\,mol\\
{n_{{H_2}}} = {n_{FeO}} = 0,4\,mol\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
hh:Al(a\,mol),Fe(b\,mol)\\
27a + 56b = 11\\
1,5a + b = 0,4\\
\Rightarrow a = 0,2;b = 0,1\\
{n_{FeC{l_2}}} = {n_{Fe}} = 0,1\,mol\\
{n_{AlC{l_3}}} = {n_{Al}} = 0,2\,mol\\
m = {m_{FeC{l_2}}} + {m_{AlC{l_3}}} = 0,1 \times 127 + 0,2 \times 133,5 = 39,4g\\
b)\\
\% {m_{Fe}} = \dfrac{{0,1 \times 56}}{{11}} \times 100\% = 50,9\% \\
\% {m_{Al}} = 100 - 50,9 = 49,1\% \\
c)\\
{n_{HCl}} = 0,2 \times 3 + 0,1 \times 2 = 0,8\,mol\\
{C_\% }HCl = \dfrac{{0,8 \times 36,5}}{{500}} \times 100\% = 5,84\%
\end{array}\)
