Em tham khảo nha :
\(\begin{array}{l}
1)\\
a)\\
{m_{S{O_2}}} = n \times M = 2 \times 64 = 128g\\
b)\\
{n_{{H_2}}} = \dfrac{V}{{22,4}} = \dfrac{{7,84}}{{22,4}} = 0,35mol\\
{m_{{H_2}}} = n \times M = 0,35 \times 2 = 0,7g\\
c)\\
{n_{FeS{O_4}}} = 0,15 \times 3 = 0,45mol\\
{m_{FeS{O_4}}} = 0,45 \times 152 = 68,4g\\
d)\\
{m_{HCl}} = \dfrac{{200 \times 7,3}}{{100}} = 14,6g\\
2)\\
a)\\
S{O_3} + {H_2}O \to {H_2}S{O_4}\\
{n_{SO3}} = \dfrac{{40}}{{80}} = 0,5mol\\
{n_{{H_2}S{O_4}}} = {n_{S{O_3}}} = 0,5mol\\
{m_{{H_2}S{O_4}}} = 0,5 \times 98 = 49g\\
b)\\
{C_{{M_{{H_2}S{O_4}}}}} = \dfrac{{0,5}}{{0,2}} = 2,5M\\
3)\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
{n_{Fe}} = \dfrac{{12}}{{56}} = \dfrac{3}{{14}}mol\\
{n_{{H_2}S{O_4}}} = \dfrac{{12,25}}{{98}} = 0,125mol\\
\dfrac{3}{{14}} > \dfrac{{0,125}}{1} \Rightarrow Fe\text{ dư}\\
{n_{{H_2}}} = {n_{{H_2}S{O_4}}} = 0,125mol\\
{V_{{H_2}}} = 0,125 \times 22,4 = 2,8l
\end{array}\)
