Đáp án:

\(\% {m_{KMn{O_4}}} = 11\% \)

Giải thích các bước giải:

 Phản ứng xảy ra:

\(2KMn{O_4}\xrightarrow{{{t^o}}}{K_2}Mn{O_4} + Mn{O_2} + {O_2}\)

Ta có:

\({n_{KMn{O_4}}} = \frac{{9,48}}{{158}} = 0,06{\text{ mol}} \to {{\text{n}}_{{O_2}{\text{ lt}}}} = \frac{1}{2}{n_{KMn{O_4}}} = 0,03{\text{ mol}}\)

\( \to {n_{{O_2}}} = 0,03.90\%  = 0,027{\text{ mol}}\)

BTKL:

\({m_X} = {m_{KMn{O_4}}} - {m_{{O_2}}} = 9,48 - 0,027.32 = 8,616{\text{ gam}}\)

\({n_{KMn{O_4}{\text{ dư}}}} = 0,06 - 0,027.2 = 0,006{\text{ mol}}\)

\( \to {m_{KMn{O_4}{\text{ dư}}}} = 0,006.158 = 0,948{\text{ gam}}\)

\( \to \% {m_{KMn{O_4}}} = \frac{{0,948}}{{8,616}} = 11\% \)

Đáp án:

$\%m_{KMnO_4\ dư}=11\%$

Giải thích các bước giải:

$n_{KMnO_4\ bđ}=\dfrac{9,48}{158}=0,06(mol)\\ H=90\%\\ \Rightarrow n_{KMnO_4\ pư}=0,06.90\%=0,054(mol)\\ \Rightarrow n_{KMnO_4\ dư}=0,06-0,054=0,006(mol)\\ 2KMnO_4\xrightarrow{t^0} K_2MnO_4+MnO_2+O_2\\ \xrightarrow{\text{Từ PTHH}}\ n_{O_2}=\dfrac{1}{2}.n_{KMnO_4\ pư}=0,027(mol)\\ \Rightarrow m_{O_2}=0,027.32=0,864(g)\\ \xrightarrow{\text{BTKL}}\ m_X=m_{KMnO_4}-m_{O_2}=9,48-0,864=8,616(g)\\ \Rightarrow \%m_{KMnO_4\ dư}=\dfrac{0,006.158}{8,616}.100\%=11\%$