Đáp án:
a) 21,9 g
b) 18g
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
nS{O_2} = \dfrac{{12,8}}{{64}} = 0,2\,mol\\
nNaOH = 0,25 \times 1 = 0,25\,mol\\
\dfrac{{nNaOH}}{{nS{O_2}}} = \dfrac{{0,25}}{{0,2}} = 1,25\\
2NaOH + S{O_2} \to N{a_2}S{O_3} + {H_2}O(1)\\
N{a_2}S{O_3} + S{O_2} + {H_2}O \to 2NaHS{O_3}(2)\\
nS{O_2}(1) = nN{a_2}S{O_3}(1) = \dfrac{{0,25}}{2} = 0,125\,mol\\
nS{O_2}(2) = nN{a_2}S{O_3}(2) = 0,2 - 0,125 = 0,075\,mol\\
nN{a_2}S{O_3} = 0,125 - 0,075 = 0,05\,mol\\
nNaS{O_3} = 0,075 \times 2 = 0,15\,mol\\
m = mNaHS{O_3} + mN{a_2}S{O_3} = 21,9g\\
b)\\
nS{O_2} = \dfrac{{4,48}}{{22,4}} = 0,2\,mol\\
nKOH = 0,1 \times 1,5 = 0,15\,mol\\
\dfrac{{nKOH}}{{nS{O_2}}} = \dfrac{{0,15}}{{0,2}} = 0,75\\
KOH + S{O_2} \to KHS{O_3}\\
nKHS{O_3} = nKOH = 0,15\,mol\\
mKHS{O_3} = 0,15 \times 120 = 18g
\end{array}\)
