Đáp án:
\( {C_{M{\text{ N}}{{\text{a}}_2}S{O_3}}} = 0,8M;{C_{M{\text{ NaOH}}}} = 0,4M\)
\(m = 29,2{\text{ gam}}\)
Giải thích các bước giải:
Ta có:
\({n_{S{O_2}}} = \frac{{4,48}}{{22,4}} = 0,2{\text{ mol;}}{{\text{n}}_{NaOH}} = 0,25.2 = 0,5{\text{ mol}}\)
\( \to \frac{{{n_{NaOH}}}}{{{n_{S{O_2}}}}} = \frac{{0,5}}{{0,2}} = 2,5 > 2\)
Vậy \(NaOH\) dư
\(2NaOH + S{O_2}\xrightarrow{{}}N{a_2}S{O_3} + {H_2}O\)
\( \to {n_{N{a_2}S{O_3}}} = {n_{S{O_2}}} = 0,2{\text{ }}mol\)
\( \to {n_{NaOH}} = 0,5 - 0,2.2 = 0,1{\text{ mol}}\)
\( \to {C_{M{\text{ N}}{{\text{a}}_2}S{O_3}}} = \frac{{0,2}}{{0,25}} = 0,8M;{C_{M{\text{ NaOH}}}} = \frac{{0,1}}{{0,25}} = 0,4M\)
\(m = {m_{N{a_2}S{O_3}}} + {m_{NaOH}} = 0,2.(23.2 + 80) + 0,1.40 = 29,2{\text{ gam}}\)
