Đáp án:
Bạn tham khảo lời giải ở dưới nhé !
Giải thích các bước giải:
Bài 1:
\(\begin{array}{l}
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
{n_{{H_2}}} = 0,3mol\\
\to {n_{Al}} = \dfrac{2}{3}{n_{{H_2}}} = 0,2mol\\
\to {m_{Al}} = 5,4g\\
\to {m_{Cu}} = 8g
\end{array}\)
\(\begin{array}{l}
a)\\
\% {m_{Al}} = \dfrac{{5,4}}{{13,4}} \times 100\% = 40,3\% \\
\% {m_{Cu}} = 100\% - 40,3\% = 59,7\%
\end{array}\)
\(\begin{array}{l}
b)\\
{n_{{H_2}S{O_4}}} = {n_{{H_2}}} = 0,3mol\\
\to C{M_{{H_2}S{O_4}}} = \dfrac{{0,3}}{{0,1}} = 3M
\end{array}\)
\(\begin{array}{l}
c)\\
{n_{A{l_2}{{(S{O_4})}_3}}} = \dfrac{1}{3}{n_{{H_2}}} = 0,1mol\\
\to C{M_{A{l_2}{{(S{O_4})}_3}}} = \dfrac{{0,1}}{{0,1}} = 1M
\end{array}\)
Bài 2:
\(\begin{array}{l}
a)\\
A{l_2}{O_3} + 6HCl \to 2AlC{l_3} + 3{H_2}O\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
{n_{{H_2}}} = 0,1mol\\
\to {n_{Zn}} = {n_{{H_2}}} = 0,1mol\\
\to {m_{Zn}} = 6,5g\\
\to {m_{A{l_2}{O_3}}} = 5,1g \to {n_{A{l_2}{O_3}}} = 0,05mol\\
\to \% {m_{Zn}} = \dfrac{{6,5}}{{11,6}} \times 100\% = 56,03\% \\
\to \% {m_{A{l_2}{O_3}}} = 100\% - 56,03\% = 43,97\%
\end{array}\)
\(\begin{array}{l}
b)\\
{n_{HCl}} = 6{n_{A{l_2}{O_3}}} + 2{n_{Zn}} = 0,5mol\\
\to {m_{HCl}} = 18,25g\\
\to {m_{{\rm{dd}}HCl}} = \dfrac{{18,25}}{{20\% }} \times 100\% = 91,25g
\end{array}\)
\(\begin{array}{l}
c)\\
Zn + C{\rm{uS}}{{\rm{O}}_4} \to ZnS{O_4} + Cu\\
{n_{Cu}} = {n_{Zn}} = 0,1mol\\
\to {m_{Cu}} = 6,4g
\end{array}\)
