Đáp án:
\(\begin{array}{l}
1)\\
{m_{Fe}} = 7,84g\\
2)\\
H = 80\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
1)\\
F{e_2}{O_3} + 3{H_2} \to 2Fe + 3{H_2}O\\
{n_{F{e_2}{O_3}}} = \dfrac{m}{M} = \dfrac{{32}}{{160}} = 0,2mol\\
{n_{{H_2}}} = \dfrac{V}{{22,4}} = \dfrac{{6,72}}{{22,4}} = 0,3mol\\
\dfrac{{0,2}}{1} > \dfrac{{0,3}}{3} \Rightarrow F{e_2}{O_3}\text{ dư}\\
{n_{Fe}} = \dfrac{2}{3}{n_{{H_2}}} = 0,2mol\\
{m_{Fe}} = n \times M = 0,2 \times 56 = 11,2g\\
{m_{Fe}\text{thu được}} = \dfrac{{11,2 \times 70}}{{100}} = 7,84g\\
2)\\
2S{O_2} + {O_2} \to 2S{O_3}\\
{n_{S{O_2}}} = \dfrac{V}{{22,4}} = \dfrac{{16,8}}{{22,4}} = 0,75mol\\
\Rightarrow {n_{S{O_3}}} = {n_{S{O_2}}} = 0,75mol\\
{m_{S{O_3}}} = n \times M = 0,75 \times 80 = 60g\\
H = \dfrac{{{m_{S{O_3}}}\text{thu được}}}{{{m_{S{O_3}}}}} \times 100\% = \dfrac{{48}}{{60}} \times 100\% = 80\%
\end{array}\)
