Cho 33,9 gam hỗn hợp bột Zn và Mg (tỉ lệ 1 : 2) t...

D

- Hướng dẫn giải

Chọn D

\(\underbrace {\overbrace {Zn}^{0,3\,mol},\,\overbrace {Mg}^{0,6\,mol}}_{{\rm{hon hop kim loai}}} + \underbrace {NaN{O_3},\,NaHS{O_4}}_{dung\,dich\,hon\,hop} \to \underbrace {\overbrace {Z{n^{2 + }}}^{0,3\,mol},\overbrace {\,M{g^{2 + }}}^{0,6\,mol},\,N{a^ + },\,N{H_4}^ + ,\,S{O_4}^{2 - }}_{{\rm{dung dich A}}} + \underbrace {\overbrace {{N_2}O}^{0,15\,mol},\,\overbrace {{H_2}}^{0,05\,mol}}_{{\rm{ hon hop B}}} + {H_2}O\)

\( \to {n_{N{H_4}^ + }} = \frac{{2{n_{Z{n^{2 + }}}} + 2{n_{M{g^{2 + }}}} - 8{n_{{N_2}O}} - 2{n_{{H_2}}}}}{8} = 0,0625\,mol \to {n_{NaN{O_3}}} = 2{n_{{N_2}O}} + {n_{N{H_4}^ + }} = 0,3625\,mol\)

\( \Rightarrow {n_{NaHS{O_4}}} = 10{n_{N{H_4}^ + }} + 10{n_{{N_2}O}} + 2{n_{{H_2}}} = 2,225\,mol \to {n_{{H_2}O}} = \frac{{{n_{NaHS{O_4}}} - 4{n_{N{H_4}^ + }} - 2{n_{{H_2}}}}}{2} = 0,9375\,mol\)

\({m_A} = {m_{kim{\rm{ loai }}}} + 85{n_{NaN{O_3}}} + 120{n_{NaHS{O_4}}} - {m_B} - 18{n_{{H_2}O}} = 308,1375\,(g)\)