1. B

2. C

3. A

4. C

5. A

Em tham khảo nha :

\(\begin{array}{l}
1)\\
B\\
S + {O_{2}} \to S{O_2}\\
{n_{{O_2}}} = \dfrac{V}{{22,4}} = \dfrac{{1,12}}{{22,4}} = 0,05mol\\
{n_S} = {n_{{O_2}}} = 0,05mol\\
{m_S} = n \times M = 0,05 \times 32 = 1,6g\\
2)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{V}{{22,4}} = \dfrac{{4,48}}{{22,4}} = 0,2mol\\
{n_{Fe}} = {n_{{H_2}}} = 0,2mol\\
{m_{Fe}} = n \times M = 0,2 \times 56 = 11,2g\\
3)\\
A\\
4P + 5{O_2} \to 2{P_2}{O_5}\\
{n_P} = \dfrac{m}{M} = \dfrac{{3,1}}{{31}} = 0,1mol\\
{n_{{O_2}}} = \dfrac{5}{4}{n_P} = 0,125mol\\
{V_{{O_2}}} = n \times 22,4 = 0,125 \times 22,4 = 2,8l\\
{n_{{P_2}{O_5}}} = \dfrac{{{n_P}}}{2} = 0,05mol\\
{m_{{P_2}{O_5}}} = n \times M = 0,05 \times 142 = 7,1g\\
4)\\
C\\
2Mg + {O_2} \to 2MgO\\
{n_{Mg}} = \dfrac{m}{M} = \dfrac{{4,8}}{{24}} = 0,2mol\\
{n_{{O_2}}} = \dfrac{{{n_{Mg}}}}{2} = 0,1mol\\
{V_{{O_2}}} = n \times 22,4 = 0,1 \times 22,4 = 2,24l\\
5)\\
D\\
2KCl{O_3} \to 2KCl + 3{O_2}\\
{n_{KCl{O_3}}} = \dfrac{m}{M} = \dfrac{{24,5}}{{122,5}} = 0,2mol\\
{n_{{O_2}}} = \dfrac{3}{2}{n_{KCl{O_3}}} = 0,3mol\\
{V_{{O_2}}} = n \times 22,4 = 0,3 \times 22,4 = 6,72l
\end{array}\)