Cho 3,48 gam bột Mg tan hết trong dung dịch hỗn hợ...

D

- Hướng dẫn giải

\({n_{Mg}} = \frac{{3,48}}{{24}} = 0,145{\rm{ mol}}\)

\(\left\{ \begin{array}{l}
{n_{{N_2}}} + {n_{{H_2}}} = \frac{{0,56}}{{22,4}} = 0,025{\rm{ mol}}\\
{\rm{28}}{{\rm{n}}_{{N_2}}} + 2{n_{{H_2}}} = 11,4.2.0,025 = 0,57g
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
{n_{{N_2}}} = 0,02{\rm{ mol}}\\
{{\rm{n}}_{{H_2}}} = 0,005{\rm{ mol}}
\end{array} \right.\)

\(\to {n_{NH_4^ + }} = \frac{{2.0,145 - 10.0,02 - 2.0,005}}{8} = 0,01{\rm{ mol}}\)

\(\to {n_{KN{O_3}}} = 2{n_{{N_2}}} + {n_{NH_4^ + }} = 0,05{\rm{ mol}}\)

m _muối \( = {m_{MgC{l_2}}} + {m_{N{H_4}Cl}} + {m_{KCl}} = 95.0,145 + 53,5.0,01 + 74,5.0,05 = 18,035g\)