Giúp em với ạ !!!!!!!!!!!!

Đáp án:

 \(\begin{array}{l}
a)\\
\% Mn{O_2} = 23,97\% \\
\% {K_2}Mn{O_4} = 54,27\% \\
\% KMn{O_4}d = 21,76\% \\
b)\\
H = 80\% 
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
2KMn{O_4} \to {K_2}Mn{O_4} + Mn{O_2} + {O_2}\\
\text{Gọi a là số mol của $KMnO_4$ phản ứng }\\
{n_{Mn{O_2}}} = \dfrac{{{n_{KMn{O_4}}}}}{2} = \dfrac{a}{2}mol\\
{n_{{K_2}Mn{O_4}}} = \dfrac{{{n_{KMn{O_4}}}}}{2} = \dfrac{a}{2}mol\\
{n_{KMn{O_4}d}} = {m_{Kmn{O_4}}} - {m_{KMn{O_4}pu}} = 79 - 158a(g)\\
{m_{KMn{O_4}d}} + {m_{Mn{O_2}}} + {m_{{K_2}Mn{O_4}}} = 72,6g\\
79 - 158a + \dfrac{{87}}{2}a + \dfrac{{197}}{2} = 72,6g\\
 \Rightarrow a = 0,4mol\\
{n_{Mn{O_2}}} = \dfrac{{{n_{KMn{O_4}}}}}{2} = 0,2mol\\
{m_{Mn{O_2}}} = n \times M = 0,2 \times 87 = 17,4g\\
{n_{{K_2}Mn{O_4}}} = \dfrac{{{n_{KMn{O_4}}}}}{2} = 0,2mol\\
{m_{{K_2}Mn{O_4}}} = n \times M = 0,2 \times 197 = 39,4g\\
\% Mn{O_2} = \dfrac{{17,4}}{{72,6}} \times 100\%  = 23,97\% \\
\% {K_2}Mn{O_4} = \dfrac{{39,4}}{{72,6}} \times 100\%  = 54,27\% \\
\% KMn{O_4}d = 100 - 23,97 - 54,27 = 21,76\% \\
b)\\
{m_{KMn{O_4}pu}} = 0,4 \times 158 = 63,2g\\
H = \dfrac{{63,2}}{{79}} \times 100\%  = 80\% 
\end{array}\)