Giải thích các bước giải:
\(\begin{array}{l}
33.\\
a)2KMn{O_4} \to {K_2}Mn{O_4} + Mn{O_2} + {O_2}\\
b)\\
{m_{KMn{O_4}}} = \dfrac{{23,7}}{{75}} \times 100 = 31,6g\\
\to {n_{KMn{O_4}}} = 0,2mol\\
\to {n_{{O_2}}} = \dfrac{1}{2}{n_{KMn{O_4}}} = 0,1mol\\
\to {V_{{O_2}}} = 2,24l\\
35.\\
BaC{O_3} \to BaO + C{O_2}\\
{n_{BaC{O_3}}} = 0,15mol\\
{n_{C{O_2}}} = 0,12mol\\
\to {n_{BaC{O_3}}} > {n_{C{O_2}}} \to {n_{BaC{O_3}}}du\\
\to {n_{BaC{O_3}(pt)}} = {n_{C{O_2}}} = 0,12mol\\
\to {n_{BaC{O_3}(dư)}} = 0,03mol\\
\to {m_{BaC{O_3}(dư)}} = 5,91g\\
{n_{BaO}} = {n_{C{O_2}}} = 0,12mol\\
\to {m_{BaO}} = 18,36g
\end{array}\)
