Đáp án:
\(\begin{array}{l}
b)\\
{m_{Fe}} = 11,2g\\
{m_{Na}} = 2,3g\\
c)\\
47,1g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2Na + 2{H_2}S{O_4} \to N{a_2}S{O_4} + S{O_2} + 2{H_2}O\\
2Fe + 6{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3S{O_2} + 6{H_2}O\\
b)\\
{n_{S{O_2}}} = \dfrac{{7,84}}{{22,4}} = 0,35\,mol\\
hh:Fe(a\,mol),Na(b\,mol)\\
56a + 23b = 13,5\\
1,5a + 0,5b = 0,35\\
\Rightarrow a = 0,2;b = 0,1\\
{m_{Fe}} = 0,2 \times 56 = 11,2g\\
{m_{Na}} = 0,1 \times 23 = 2,3g\\
c)\\
{n_{N{a_2}S{O_4}}} = \dfrac{{{n_{Na}}}}{2} = \dfrac{{0,1}}{2} = 0,05\,mol\\
{n_{F{e_2}{{(S{O_4})}_3}}} = \dfrac{{{n_{Fe}}}}{2} = \dfrac{{0,2}}{2} = 0,1\,mol\\
m = {m_{N{a_2}S{O_4}}} + {m_{F{e_2}{{(S{O_4})}_3}}} = 0,05 \times 142 + 0,1 \times 400 = 47,1g
\end{array}\)