Đáp án:

a, \({V_{{O_2}}}thực= \dfrac{{6,72 \times 75}}{{100}} = 5,04l\)

b, \({m_{KMn{O_4}}}cần= \dfrac{{474 \times 100}}{{80}} = 592,5g\)

Giải thích các bước giải:

\(\begin{array}{l}
a,\\
2KMn{O_4} \to {K_2}Mn{O_4} + Mn{O_2} + {O_2}\\
{n_{KMn{O_4}}} = 0,6mol\\
 \to {n_{{O_2}}} = \dfrac{1}{2}{n_{KMn{O_4}}} = 0,3mol\\
 \to {V_{{O_2}}} = 6,72l\\
 \to {V_{{O_2}}}thực= \dfrac{{6,72 \times 75}}{{100}} = 5,04l\\
b,\\
2KMn{O_4} \to {K_2}Mn{O_4} + Mn{O_2} + {O_2}\\
{n_{{O_2}}} = 1,5mol\\
 \to {n_{KMn{O_4}}} = 2{n_{{O_2}}} = 3mol\\
 \to {m_{KMn{O_4}}} = 474g\\
 \to {m_{KMn{O_4}}}cần= \dfrac{{474 \times 100}}{{80}} = 592,5g
\end{array}\)

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