Đáp án:
\(\begin{array}{l}
a)\\
m = 2,178g\\
{m_{N{a_2}Zn{O_2}}} = 3,146g\\
b)\\
{V_{HCl}} = 20,95ml
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Zn{(OH)_2} + 2NaOH \to N{a_2}Zn{O_2} + 2{H_2}O\\
{n_{NaOH}} = 0,022 \times 2 = 0,044\,mol\\
{n_{Zn{{(OH)}_2}}} = {n_{N{a_2}Zn{O_2}}} = \dfrac{{0,044}}{2} = 0,022\,mol\\
m = {m_{Zn{{(OH)}_2}}} = 0,022 \times 99 = 2,178g\\
{m_{N{a_2}Zn{O_2}}} = 0,022 \times 143 = 3,146g\\
b)\\
Zn{(OH)_2} + 2HCl \to ZnC{l_2} + 2{H_2}O\\
{n_{HCl}} = 2{n_{Zn{{(OH)}_2}}} = 0,022 \times 2 = 0,044\,mol\\
{m_{{\rm{dd}}HCl}} = \dfrac{{0,044 \times 36,5}}{{7,3\% }} = 22g\\
{V_{HCl}} = \dfrac{{22}}{{1,05}} = 20,95ml
\end{array}\)
