Đáp án:
\(a = 9,75{\text{ gam}}\)
\({m_{ZnC{l_2}}} = 20,4{\text{ gam}}\)
\({C_{M{\text{ ZnC}}{{\text{l}}_2}}} = 0,75M\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(Zn + 2HCl\xrightarrow{{}}ZnC{l_2} + {H_2}\)
Ta có:
\({n_{{H_2}}} = \frac{{3,36}}{{22,4}} = 0,15{\text{ mol = }}{{\text{n}}_{Zn}} = {n_{ZnC{l_2}}}\)
\( \to a = {m_{Zn}} = 0,15.65 = 9,75{\text{ gam}}\)
\({m_{ZnC{l_2}}} = 0,15.(65 + 35,5.2) = 20,4{\text{ gam}}\)
\({V_{dd}} = {V_{dd{\text{ HCl}}}} = 200{\text{ ml = 0}}{\text{,2 lít}}\)
\({C_{M{\text{ ZnC}}{{\text{l}}_2}}} = \frac{{0,15}}{{0,2}} = 0,75M\)