Cho 31,15 gam hỗn hợp bột Zn và Mg (tỷ lệ mol 1 :...

A

- Hướng dẫn giải

\(\left\{ \begin{gathered} {n_{Mg}} = 0,35(mol) \hfill \\ {n_{Zn}} = 0,35(mol) \hfill \\ \end{gathered} \right. \to {n_e} = 1,4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{n_B}\, = 0,2(mol)\left\{ \begin{gathered} {N_2}O:0,1(mol) \hfill \\ {H_2}:0,1(mol) \hfill \\ \end{gathered} \right.\)

\(\xrightarrow{{BTE}}{n_{NH_4^ + }} = \frac{{1,4 - 0,1.8 - 0,1.2}}{8} = 0,05(mol)\)

\(\to {\text{dd}}\,\,A\left\{ \begin{gathered} M{g^{2 + }}:0,35 \hfill \\ Z{n^{2 + }}:0,35 \hfill \\ NH_4^ + :0,05 \hfill \\ \xrightarrow{{BTNT.Nito}}N{a^ + }:0,1.2 + 0,05 + a = 0,25 + a \hfill \\ \xrightarrow{{BTDT}}SO_4^{2 - }:a \hfill \\ \end{gathered} \right.\)

\(\xrightarrow{{BTKL}}m = 240,1(gam)\)