Em tham khảo nha :

\(\begin{array}{l}
1)\\
a)\\
ZnO + 2HCl \to ZnC{l_2} + {H_2}O\\
{n_{ZnO}} = \dfrac{{1,62}}{{81}} = 0,02mol\\
{n_{HCl}} = 2{n_{ZnO}} = 0,04mol\\
{V_{HCl}} = \dfrac{{0,04}}{{0,4}} = 0,1l\\
b)\\
{n_{{H_2}}} = {n_{ZnO}} = 0,02mol\\
{V_{{H_2}}} = 0,02 \times 22,4 = 0,448l\\
c)\\
{n_{ZnC{l_2}}} = {n_{ZnO}} = 0,02mol\\
{C_{{M_{ZnC{l_2}}}}} = \dfrac{{0,02}}{{0,1}} = 0,2M\\
2)\\
2NaOH + {H_2}S{O_4} \to N{a_2}S{O_4} + 2{H_2}O\\
{n_{NaOH}} = 0,2 \times 0,2 = 0,04mol\\
{n_{{H_2}S{O_4}}} = 0,3 \times 0,1 = 0,03mol\\
\dfrac{{0,04}}{2} < \dfrac{{0,03}}{1} \Rightarrow {H_2}S{O_4}\text{ dư}\\
{n_{N{a_2}S{O_4}}} = \dfrac{{{n_{NaOH}}}}{2} = 0,02mol\\
{n_{{H_2}S{O_4}d}} = {n_{{H_2}S{O_4}}} - \frac{{{n_{NaOH}}}}{2} = 0,01mol\\
{C_{{M_{N{a_2}S{O_4}}}}} = \dfrac{{0,02}}{{0,2 + 0,3}} = 0,04M\\
{C_{{M_{{H_2}S{O_4}}}}} = \dfrac{{0,01}}{{0,5}} = 0,02M\\
3)\\
RO + 2HCl \to RC{l_2} + {H_2}O\\
{n_{HCl}} = 0,8 \times 0,5 = 0,4mol\\
{n_{RO}} = \dfrac{{{n_{HCl}}}}{2} = 0,2mol\\
{M_{RO}} = \dfrac{8}{{0,2}} = 40dvC\\
 \Rightarrow {M_R} = 40 - 16 = 24dvC\\
R:Magie(Mg)\\
CTHH:MgO
\end{array}\)

Đáp án:

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