Giải thích các bước giải:
1,
\(\begin{array}{l}
2Mg + {O_2} \to 2MgO\\
{n_{Mg}} = {n_{MgO}} = 2{n_{{O_2}}}\\
4Al + 3{O_2} \to 2A{l_2}{O_3}\\
{n_{Al}} = \dfrac{4}{3}{n_{{O_2}}},{n_{A{l_2}{O_3}}} = \dfrac{2}{3}{n_{{O_2}}}\\
{C_2}{H_6}O + 3{O_2} \to 2C{O_2} + 3{H_2}O\\
{n_{{C_2}{H_6}O}} = \dfrac{1}{3}{n_{{O_2}}}\\
{n_{C{O_2}}} = \dfrac{2}{3}{n_{{O_2}}}\\
{n_{{H_2}O}} = {n_{{O_2}}}
\end{array}\)
2,
\(\begin{array}{l}
{C_2}{H_6}O + 3{O_2} \to 2C{O_2} + 3{H_2}O\\
{n_{{C_2}{H_6}O}} = 0,2mol\\
{n_{C{O_2}}} = 2{n_{{C_2}{H_6}O}} = 0,4mol\\
{n_{{H_2}O}} = 3{n_{{C_2}{H_6}O}} = 0,6mol
\end{array}\)
3,
\(\begin{array}{l}
4Al + 3{O_2} \to 2A{l_2}{O_3}\\
{n_{{O_2}}} = 0,45mol\\
{n_{Al}} = \dfrac{4}{3}{n_{{O_2}}} = 0,6mol \to m = 16,2g\\
{n_{A{l_2}{O_3}}} = \dfrac{2}{3}{n_{{O_2}}} = 0,3mol \to m = 30,6g
\end{array}\)
