Em tham khảo nha:
\(\begin{array}{l}
3)\\
a)\\
S{O_3} + {H_2}O \to {H_2}S{O_4}\\
S{O_3}:{H_2}O:{H_2}S{O_4} = 1:1:1\\
b)\\
C{O_2} + {H_2}O \to {H_2}C{O_3}\\
C{O_2}:{H_2}O:{H_2}C{O_3} = 1:1:1\\
c)\\
HCl + NaOH \to NaCl + {H_2}O\\
HCl:NaOH:NaCl:{H_2}O = 1:1:1:1\\
d)\\
MgO + 2HN{O_3} \to Mg{(N{O_3})_2} + {H_2}O\\
MgO:HN{O_3}:Mg{(N{O_3})_2}:{H_2}O = 1:2:1:1\\
e)\\
2Al + 3CuO \to A{l_2}{O_3} + 3Cu\\
Al:CuO:A{l_2}{O_3}:Cu = 2:3:1:3\\
2.1)\\
a)\\
{n_{Fe}} = \dfrac{m}{M} = \dfrac{{28}}{{56}} = 0,5\,mol\\
b)\\
{n_{{H_2}O}} = n \times M = 0,25 \times 18 = 4,5g\\
c)\\
{n_{{N_2}O}} = \dfrac{m}{M} = \dfrac{{3,3}}{{44}} = 0,075\,mol\\
{V_{{N_2}O}} = n \times 22,4 = 0,075 \times 22,4 = 1,68l\\
d)\\
{n_{C{H_4}}} = \dfrac{V}{{22,4}} = \dfrac{{5,6}}{{22,4}} = 0,25\,mol\\
{m_{C{H_4}}} = n \times M = 0,25 \times 16 = 4g\\
2.2)\\
a)\\
{n_{{H_2}S{O_4}}} = \dfrac{m}{M} = \dfrac{{4,9}}{{98}} = 0,05\,mol\\
b)\\
{m_{N{a_2}C{O_3}}} = n \times M = 0,2 \times 106 = 21,2g\\
c)\\
{n_{C{O_2}}} = \dfrac{m}{M} = \dfrac{{95,48}}{{44}} = 2,17\,mol\\
{V_{C{O_2}}} = n \times 22,4 = 2,17 \times 22,4 = 48,608l\\
d)\\
{V_{C{O_2}}} = n \times 22,4 = 0,25 \times 22,4 = 5,6l\\
2.3)\\
a)\\
{V_{{O_2}}} = n \times 22,4 = 0,2 \times 22,4 = 4,48l\\
b)\\
{n_{C{l_2}}} = \dfrac{V}{{22,4}} = \dfrac{{2,8}}{{22,4}} = 0,125\,mol
\end{array}\)
