Đáp án:
\(\begin{array}{l}
b)\\
{V_{{H_2}}} = 11,2l\\
c)\\
\% {m_{Mg}} = 75\% \\
\% {m_{MgO}} = 25\% \\
d)\\
{C_M}MgS{O_4} = 3M
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Mg + {H_2}S{O_4} \to MgS{O_4} + {H_2}\\
MgO + {H_2}S{O_4} \to MgS{O_4} + {H_2}O\\
b)\\
{n_{MgS{O_4}}} = \dfrac{{72}}{{120}} = 0,6\,mol\\
hh:Mg(a\,mol),MgO(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 0,6\\
24a + 40b = 16
\end{array} \right.\\
\Rightarrow a = 0,5;b = 0,1\\
{n_{{H_2}}} = {n_{Mg}} = 0,5\,mol\\
{V_{{H_2}}} = 0,5 \times 22,4 = 11,2l\\
c)\\
\% {m_{Mg}} = \dfrac{{0,5 \times 24}}{{16}} \times 100\% = 75\% \\
\% {m_{MgO}} = 100 - 75 = 25\% \\
d)\\
{n_{{H_2}S{O_4}}} = 0,2 \times 2 = 0,4\,mol\\
{C_M}MgS{O_4} = \dfrac{{0,6}}{{0,2}} = 3M
\end{array}\)