Đáp án:
\(\begin{array}{l}
a)\\
\% {m_{Al}} = 72,97\% \\
\% {m_{Ca}} = 27,03\% \\
b)\\
{m_{{\rm{dd}}HCl}} = 350g\\
c)\\
{C_M}HCl = 2,24M\\
d)\\
{C_\% }AlC{l_3} = 7,49\% \\
{C_\% }CaC{l_2} = 1,56\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
Ca + 2HCl \to CaC{l_2} + {H_2}\\
hh:Al(a\,mol),Ca(b\,mol)\\
{n_{{H_2}}} = \dfrac{{7,84}}{{22,4}} = 0,35\,mol\\
\left\{ \begin{array}{l}
27a + 40b = 7,4\\
1,5a + b = 0,35
\end{array} \right.\\
\Rightarrow a = 0,2;b = 0,05\\
\% {m_{Al}} = \dfrac{{0,2 \times 27}}{{7,4}} \times 100\% = 72,97\% \\
\% {m_{Ca}} = 100 - 72,97 = 27,03\% \\
b)\\
{n_{HCl}} = 2{n_{{H_2}}} = 0,7\,mol\\
{m_{{\rm{dd}}HCl}} = \dfrac{{0,7 \times 36,5}}{{7,3\% }} = 350g\\
c)\\
{V_{{\rm{dd}}HCl}} = \dfrac{{350}}{{1,12}} = 312,5\,ml = 0,3125l\\
{C_M}HCl = \dfrac{{0,7}}{{0,3125}} = 2,24M\\
d)\\
{m_{{\rm{dd}}spu}} = 7,4 + 350 - 0,35 \times 2 = 356,7g\\
{C_\% }AlC{l_3} = \dfrac{{0,2 \times 133,5}}{{356,7}} \times 100\% = 7,49\% \\
{C_\% }CaC{l_2} = \dfrac{{0,05 \times 111}}{{356,7}} \times 100\% = 1,56\%
\end{array}\)
