Đáp án:
\(\to {m_{KMn{O_4}}} = 31,6{\text{ gam}}\)
Giải thích các bước giải:
\(2Mg + {O_2}\xrightarrow{{}}2MgO\)
\(2KMn{O_4}\xrightarrow{{}}{K_2}Mn{O_4} + Mn{O_2} + {O_2}\)
\({n_{Mg}} = \frac{{4,8}}{{24}} = 0,2{\text{ mol}} \to {{\text{n}}_{{O_2}}} = \frac{1}{2}{n_{Mg}} = 0,1{\text{ mol}} \to {{\text{n}}_{KMn{O_4}}} = 2{n_{{O_2}}} = 0,2{\text{ }}mol\)
\(\to {m_{KMn{O_4}}} = 0,2.(39 + 55 + 16.3) = 31,6{\text{ gam}}\)