Em tham khảo nha:
\(\begin{array}{l}
1)\\
a)\,{m_{{O_2}}} = 3 \times 32 = 96g\\
b)\,{n_{NaOH}} = \dfrac{{20}}{{40}} = 0,5\,mol\\
c)\,{V_{C{O_2}}} = 2 \times 22,4 = 44,8l\\
d)\,\\
{d_1}:\\
{C_\% }{H_2}S{O_4} = \dfrac{{30}}{{150}} \times 100\% = 20\% \\
{d_2}:\\
{C_\% }KCl = \dfrac{{15}}{{50}} \times 100\% = 30\% \\
e)\\
{C_M}HCl = \dfrac{{0,1}}{{0,2}} = 0,5M\\
2)\\
a)\,C + {O_2} \xrightarrow{t^0} C{O_2}\\
b)\,2{H_2} + {O_2} \xrightarrow{t^0} 2{H_2}O\\
c)\,F{e_2}{O_3} + 3{H_2} \xrightarrow{t^0} 2Fe + 3{H_2}O\\
d)\,CaO + {H_2}O \to Ca{(OH)_2}\\
e)\,2Na + 2{H_2}O \to 2NaOH + {H_2}\\
f)\,S{O_2} + {H_2}O \to {H_2}S{O_3}\\
3)\\
a)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
b)\\
{n_{Fe}} = \dfrac{{5,6}}{{56}} = 0,1\,mol\\
{n_{HCl}} = 2{n_{Fe}} = 0,2\,mol\\
{C_M}HCl = \dfrac{{0,2}}{{0,2}} = 1M\\
{n_{{H_2}}} = {n_{Fe}} = 0,1\,mol\\
{V_{{H_2}}} = 0,1 \times 22,4 = 2,24l
\end{array}\)
