Đáp án:
\(\begin{array}{l}
b)\\
\% Al = 19,42\% \\
\% Fe = 80,58\% \\
c)\\
{m_{{\rm{dd}}HCl}} = 700g\\
d)\\
C{\% _{AlC{l_3}}} = 1,87\% \\
C{\% _{FeC{l_2}}} = 3,56\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
b)\\
{n_{{H_2}}} = \dfrac{V}{{22,4}} = \dfrac{{7,84}}{{22,4}} = 0,35mol\\
hh:Al(a\,mol),Fe(b\,mol)\\
\dfrac{3}{2}a + b = 0,35(1)\\
27a + 56b = 13,9(2)\\
\text{Từ (1) và (2)}\Rightarrow a = 0,1;b = 0,2\\
{m_{Al}} = n \times M = 0,1 \times 27 = 2,7g\\
{m_{Fe}} = 13,9 - 2,7 = 11,2g\\
\% Al = \dfrac{{2,7}}{{13,9}} \times 100\% = 19,42\% \\
\% Fe = 100 - 19,42 = 80,58\% \\
c)\\
{n_{HCl}} = 3{n_{Al}} + 2{n_{Fe}} = 3 \times 0,1 + 2 \times 0,2 = 0,7mol\\
{m_{HCl}} = n \times M = 0,7 \times 36,5 = 25,55g\\
{m_{{\rm{dd}}HCl}} = \dfrac{{25,55 \times 100}}{{3,65}} = 700g\\
d)\\
{n_{AlC{l_3}}} = {n_{Al}} = 0,1mol\\
{m_{AlC{l_3}}} = n \times M = 0,1 \times 133,5 = 13,35g\\
{n_{FeC{l_2}}} = {n_{Fe}} = 0,2mol\\
{m_{FeC{l_2}}} = n \times M = 0,2 \times 127 = 25,4g\\
{m_{{\rm{dd}}spu}} = 13,9 + 700 - 0,35 \times 2 = 713,2g\\
C{\% _{AlC{l_3}}} = \dfrac{{13,35}}{{713,2}} \times 100\% = 1,87\% \\
C{\% _{FeC{l_2}}} = \dfrac{{25,4}}{{713,2}} \times 100\% = 3,56\%
\end{array}\)
