Đáp án:
\(\begin{array}{l}
a)\\
{V_{{H_2}}} = 2,24l\\
b)\\
{C_M}MgS{O_4} = 1M\\
{C_M}{H_2}S{O_4} = 5M
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Mg + {H_2}S{O_4} \to MgS{O_4} + {H_2}\\
{n_{Mg}} = \dfrac{{2,4}}{{24}} = 0,1\,mol\\
{n_{{H_2}S{O_4}}} = 0,1 \times 6 = 0,6\,mol\\
{n_{Mg}} < {n_{{H_2}S{O_4}}} \Rightarrow {H_2}S{O_4} \text{ dư } \\
{n_{{H_2}}} = {n_{Mg}} = 0,1\,mol\\
{V_{{H_2}}} = 0,1 \times 22,4 = 2,24l\\
b)\\
{n_{MgS{O_4}}} = {n_{Mg}} = 0,1\,mol\\
{n_{{H_2}S{O_4}}} \text{ dư }= 0,6 - 0,1 = 0,5\,mol\\
{C_M}MgS{O_4} = \dfrac{{0,1}}{{0,1}} = 1M\\
{C_M}{H_2}S{O_4} = \dfrac{{0,5}}{{0,1}} = 5M\\
{m_{MgS{O_4}}} \text{ dư }= 0,1 \times 120 = 12g
\end{array}\)