Đáp án:
b) 8,1g và 6,5g
c) 27,2g
d) 14,6g
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Zn + 2HCl \to ZnC{l_2} + {H_2}(1)\\
ZnO + 2HCl \to ZnC{l_2} + {H_2}(2)\\
b)\\
{n_{{H_2}}} = \dfrac{V}{{22,4}} = \dfrac{{2,24}}{{22,4}} = 0,1\,mol\\
Theo\,pt(1):{n_{Zn}} = {n_{{H_2}}} = 0,1\,mol\\
{m_{Zn}} = n \times M = 0,1 \times 65 = 6,5g\\
{m_{ZnO}} = 14,6 - 6,5 = 8,1g\\
c)\\
{n_{ZnO}} = \dfrac{m}{M} = \dfrac{{8,1}}{{81}} = 0,1\,mol\\
Theo\,pt(1):{n_{ZnC{l_2}}} = {n_{Zn}} = 0,1\,mol\\
Theo\,pt(2):{n_{ZnC{l_2}}} = {n_{ZnO}} = 0,1\,mol\\
{n_{ZnC{l_2}}} = 0,1 + 0,1 = 0,2\,mol\\
{m_{ZnC{l_2}}} = n \times M = 0,2 \times 136 = 27,2g\\
d)\\
Theo\,pt(1):{n_{HCl}} = 2{n_{Zn}} = 0,2\,mol\\
Theo\,pt(2):{n_{HCl}} = {n_{ZnO}} = 0,2\,mol\\
{n_{HCl}} = 0,2 + 0,2 = 0,4\,mol\\
{m_{HCl}} = n \times M = 0,4 \times 36,5 = 14,6g
\end{array}\)
