Đáp án:

2kmno4--to->k2mno4+mno2+o2(1)

 n o2=3,36/22,4=0,15(mol)

n kmno4=63,2/158=0,4(mol)

theo(1): 0,4/2>0,15/1

=>kmno4 dư

n kmno4 pư =2.n o2=2.0,15=0,3(mol)

=>n kmno4 dư=0,4-0,3=0,1(mol)

chất rắn A gồm k2mno4,mno2,kmno4 dư

theo(1):n k2mno4=n mno2=n o2=0,,15(mol)

kmno4+hcl--->kcl+hmno4(2)

k2mno4+4hcl--->2kcl+mncl2+2h2o+o2(3)

mno2+2hcl--->mncl2+h2o(4)

sau pư (2)(3)(4) chỉ có khí o2 thoát ra ở pư (3)

theo(3):n o2=n k2mno4=0,15(mol)

=>V o2=0,15.22,4=3,36(mol)

Em tham khảo nha :

\(\begin{array}{l}
2KMn{O_4} \to {K_2}Mn{O_4} + Mn{O_2} + {O_2}\\
2KMn{O_4} + 16HCl \to 5C{l_2} + 2KCl + 2MnC{l_2} + 8{H_2}O\\
Mn{O_2} + 4HCl \to MnC{l_2} + C{l_2} + 2{H_2}O\\
{K_2}Mn{O_4} + 8HCl \to 2C{l_2} + 2KCl + MnC{l_2} + 4{H_2}O\\
{n_{{O_2}}} = \dfrac{{3,36}}{{22,4}} = 0,15mol\\
{n_{{K_2}Mn{O_4}}} = {n_{Mn{O_2}}} = {n_{{O_2}}} = 0,15mol\\
{n_{KMn{O_4}pu}} = 2{n_{{O_2}}} = 0,3mol\\
{n_{KMn{O_4}}} = \dfrac{{63,2}}{{158}} = 0,4mol\\
{n_{KMn{O_4}d}} = 0,4 - 0,3 = 0,1mol\\
{n_{C{l_2}}} = \dfrac{5}{2}{n_{KMn{O_4}d}} + {n_{Mn{O_2}}} + 2{n_{{K_2}Mn{O_4}}}\\
{n_{C{l_2}}} = 0,7mol\\
{V_{C{l_2}}} = 0,7 \times 22,4 = 15,68l
\end{array}\)