Em tham khảo nha :
\(\begin{array}{l}
3)\\
a)\\
S + {O_2} \xrightarrow{t^0} S{O_2}\\
b)\\
\text{Theo định luật bảo toàn khối lượng ta có :}\\
{m_S} + {m_{{O_2}}} = {m_{S{O_2}}}\\
\Rightarrow {m_{S{O_2}}} = 32 + 32 = 64g\\
4)\\
a)\\
F{e_3}{O_4} + 4{H_2} \to 3Fe + 4{H_2}O\\
b)\\
{n_{Fe}} = \dfrac{m}{M} = \dfrac{{67,2}}{{56}} = 1,2mol\\
{n_{F{e_3}{O_4}}} = \dfrac{{{n_{Fe}}}}{3} = 0,4mol\\
{m_{F{e_3}{O_4}}} = n \times M = 0,4 \times 232 = 92,8g\\
c)\\
{n_{{H_2}}} = 4{n_{F{e_3}{O_4}}} = 1,6mol\\
{V_{{H_2}}} = n \times 22,4 = 1,6 \times 22,4 = 35,84l\\
d)\\
2{H_2} + {O_2} \to 2{H_2}O\\
{n_{{H_2}O}} = {n_{{H_2}}} = 1,6mol\\
{m_{{H_2}O}} = n \times m = 1,6 \times 18 = 28,8g\\
7)\\
CTHH\,A:X{H_2}\\
{M_A} = 2{M_{N{H_3}}} = 34dvC\\
\Rightarrow {M_X} + 2{M_H} = 34\\
\Rightarrow {M_X} = 32dvC\\
\Rightarrow X:\text{Lưu huỳnh}(S)\\
CTHH:{H_2}S
\end{array}\)
\(\begin{array}{l}
6)\\
CaC{O_3} \xrightarrow{t^0} CaO + C{O_2}\\
{n_{CaO}} = \dfrac{{7200}}{{56}} = \dfrac{{900}}{7}\,mol\\
{n_{C{O_2}}} = {n_{CaO}} = \dfrac{{900}}{7}\,mol\\
{V_{C{O_2}}} = \dfrac{{900}}{7} \times 22,4 = 2880l = 2,88{m^3}
\end{array}\)
