Đáp án:
\(\begin{array}{l}
{V_{{H_2}}} = 3,36l\\
{C_\% }{H_2}S{O_4} = 4,92\% \\
{C_\% }MgS{O_4} = 9,03\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
Mg + {H_2}S{O_4} \to MgS{O_4} + {H_2}\\
{n_{Mg}} = \dfrac{{3,6}}{{24}} = 0,15\,mol\\
{n_{{H_2}S{O_4}}} = \dfrac{{196 \times 12,5\% }}{{98}} = 0,25\,mol\\
{n_{Mg}} < {n_{{H_2}S{O_4}}} \Rightarrow \text{ $H_2SO_4$ dư tính theo Mg} \\
{n_{{H_2}}} = {n_{Mg}} = 0,15\,mol\\
{V_{{H_2}}} = 0,15 \times 22,4 = 3,36l\\
{m_{{\rm{dd}}spu}} = 3,6 + 196 + 0,15 \times 2 = 199,3g\\
{n_{{H_2}S{O_4}}} \text{ dư }= 0,25 - 0,15 = 0,1\,mol\\
{n_{MgS{O_4}}} = {n_{Mg}} = 0,15\,mol\\
{C_\% }{H_2}S{O_4} \text{ dư }= \dfrac{{0,1 \times 98}}{{199,3}} \times 100\% = 4,92\% \\
{C_\% }MgS{O_4} = \dfrac{{0,15 \times 120}}{{199,3}} \times 100\% = 9,03\%
\end{array}\)
