Đáp án:
\( \% {m_{Zn}} = 61,9\% ; \% {m_{Cu}} = 38,1\% \)
\(m_{dd H_2SO_4}=98 gam\)
Giải thích các bước giải:
\(Cu\) không phản ứng với \(H_2SO_4\) loãng.
Phản ứng xảy ra:
\(Zn + {H_2}S{O_4}\xrightarrow{{}}ZnS{O_4} + {H_2}\)
Ta có:
\({n_{{H_2}}} = \frac{{2,24}}{{22,4}} = 0,1{\text{ mol = }}{{\text{n}}_{Zn}}\)
\( \to {m_{Zn}} = 0,1.65 = 6,5{\text{ gam}}\)
\( \to \% {m_{Zn}} = \frac{{6,5}}{{10,5}} = 61,9\% \to \% {m_{Cu}} = 38,1\% \)
\({n_{{H_2}S{O_4}}} = {n_{Zn}} = 0,1{\text{ mol}}\)
\( \to {m_{{H_2}S{O_4}}} = 0,1.98 = 9,8{\text{ gam}}\)
\( \to {m_{dd{\text{ }}{{\text{H}}_2}S{O_4}}} = \frac{{9,8}}{{10\% }} = 98{\text{ gam}}\)
BTKL:
\({m_{Zn}} + {m_{dd\;{{\text{H}}_2}S{O_4}}} = {m_{dd}} + {m_{{H_2}}}\)
\( \to 0,1.65 + 98 = {m_{dd}} + 0,1.2 \to {m_{dd}} = 104,3{\text{ gam}}\)
\({n_{ZnS{O_4}}} = {n_{Zn}} = 0,1 \to {m_{ZnS{O_4}}} = 0,1.(65 + 96) = 16,1{\text{ gam}}\)
\( \to C{\% _{ZnS{O_4}}} = \frac{{16,1}}{{104,3}} = 15,436\% \)
