Đáp án:
\(\% {m_{Zn}} = 61,6\% ; \% {m_{ZnO}} = 38,4\% \)
\( {m_{ZnC{l_2}}} = 40,8{\text{ gam}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(Zn + 2HCl\xrightarrow{{}}ZnC{l_2} + {H_2}\)
\(ZnO + 2HCl\xrightarrow{{}}ZnC{l_2} + {H_2}O\)
Ta có:
\({n_{Zn}} = {n_{{H_2}}} = \frac{{4,48}}{{22,4}} = 0,2{\text{ mol}}\)
\( \to {m_{Zn}} = 0,2.65 = 13{\text{ gam}}\)
\( \to {m_{ZnO}} = 21,1 - 13 = 8,1{\text{ gam}}\)
\( \to {n_{ZnO}} = \frac{{8,1}}{{65 + 16}} = 0,1{\text{ mol}}\)
\(\% {m_{Zn}} = \frac{{13}}{{21,1}} = 61,6\% \to \% {m_{ZnO}} = 38,4\% \)
\({n_{ZnC{l_2}}} = {n_{Zn}} + {n_{ZnO}} = 0,2 + 0,1 = 0,3{\text{ mol}}\)
\( \to {m_{ZnC{l_2}}} = 0,3.(65 + 35,5.2) = 40,8{\text{ gam}}\)
