Giải thích các bước giải:
1,
\(\begin{array}{l}
2KMn{O_4} \to {K_2}Mn{O_4} + Mn{O_2} + {O_2}\\
{n_{KMn{O_4}}} = 0,2mol\\
{n_{{O_2}}} = \dfrac{1}{2}{n_{KMn{O_4}}} = 0,1mol \to {V_{{O_2}}} = 2,24l\\
3Fe + 2{O_2} \to F{e_3}{O_4}\\
{n_{Fe}} = 0,2mol\\
\dfrac{{{n_{Fe}}}}{3} > \dfrac{{{n_{{O_2}}}}}{2}\\
{n_{F{e_3}{O_4}}} = \dfrac{1}{2}{n_{{O_2}}} = 0,05mol \to m = 11,6g
\end{array}\)
2,
\(\begin{array}{l}
{C_2}{H_6} + \dfrac{7}{2}{O_2} \to 2C{O_2} + 3{H_2}O\\
{n_{{C_2}{H_6}}} = 0,6mol\\
{n_{{O_2}}} = \dfrac{7}{2}{n_{{C_2}{H_6}}} = 2,1mol \to {V_{{O_2}}} = 47,04l\\
{n_{C{O_2}}} = 2{n_{{C_2}{H_6}}} = 1,2mol \to {V_{C{O_2}}} = 26,88l\\
{n_{{H_2}O}} = 3{n_{{C_2}{H_6}}} = 1,8mol \to {m_{{H_2}O}} = 32,4g
\end{array}\)