Đáp án:
a) 39,39% và 60,61%
b) 112 ml
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
CaC{O_3} + 2HCl \to CaC{l_2} + C{O_2} + {H_2}O\\
{V_{C{O_2}}} = {V_{\text{ giảm }}} = 8,96l\\
{n_{C{O_2}}} = \dfrac{{8,96}}{{22,4}} = 0,4\,mol\\
{n_{{H_2}}} = \dfrac{{17,92 - 8,96}}{{22,4}} = 0,4\,mol\\
{n_{Zn}} = {n_{{H_2}}} = 0,4\,mol\\
{n_{CaC{O_3}}} = {n_{C{O_2}}} = 0,4\,mol\\
\% {m_{Zn}} = \dfrac{{0,4 \times 65}}{{0,4 \times 65 + 0,4 \times 100}} \times 100\% = 39,39\% \\
\% {m_{CaC{O_3}}} = 100 - 39,39 = 60,61\% \\
b)\\
2KOH + C{O_2} \to {K_2}C{O_3} + {H_2}O\\
{n_{KOH}} = 2{n_{C{O_2}}} = 0,8\,mol\\
{m_{{\rm{dd}}KOH}} = \dfrac{{0,8 \times 56}}{{32\% }} = 140g\\
{V_{{\rm{dd}}KOH}} = \dfrac{{140}}{{1,25}} = 112ml
\end{array}\)