Đáp án:
\( {m_{dd\;{\text{A}}}} = 100,48{\text{ gam}}\)
\( \% {m_{Al}} = 54,6\% ; \% {m_{Zn}} = 45,4\% \)
Giải thích các bước giải:
Phản ứng xảy ra:
\(2Al + 3{H_2}S{O_4}\xrightarrow{{}}A{l_2}{(S{O_4})_3} + 3{H_2}\)
\(Zn + {H_2}S{O_4}\xrightarrow{{}}ZnS{O_4} + {H_2}\)
Gọi số mol \(Al;Zn\) lần lượt là \(x;y\)
\( \to 27x + 65y = 2,68{\text{ gam}}\)
\({n_{{H_2}}} = \frac{{2,24}}{{22,4}} = 0,1{\text{ mol = 1}}{\text{,5x + y = }}{{\text{n}}_{{H_2}S{O_4}}}\)
Giải được:
\(x=0,05418;y=0,01873\)
\({m_{{H_2}S{O_4}}} = 0,1.98 = 9,8{\text{ gam}}\)
\( \to {m_{dd{\text{ }}{{\text{H}}_2}S{O_4}}} = \frac{{9,8}}{{10\% }} = 98{\text{ gam}}\)
BTKL:
\({m_{kl}} + {m_{dd{\text{ }}{{\text{H}}_2}S{O_4}}} = {m_{dd\;{\text{A}}}} + {m_{{H_2}}}\)
\( \to 2,68 + 98 = {m_{dd\;{\text{A}}}} + 0,1.2 \to {m_{dd\;{\text{A}}}} = 100,48{\text{ gam}}\)
\({m_{Al}} = 0,05418.27 = 1,46286{\text{ gam}}\)
\( \to \% {m_{Al}} = \frac{{1,46286}}{{2,68}} = 54,6\% \to \% {m_{Zn}} = 45,4\% \)
