Gọi $n_{KClO_3} = a(mol) ; n_{KMnO_4} = b(mol) \Rightarrow 122,5a + 158b = 87,7(1)$
TH1 :
$2KClO_3 \xrightarrow{t^o} 2KCl + 3O_2$
$2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2$
$m_{O_2} = 87,7 - 71,7 = 16(gam)$
$\Rightarrow n_{O_2} = 1,5a + 0,5b = \dfrac{16}{32} = 0,5(2)$
Từ (1)(2) suy ra a = 0,2 ; b = 0,4
$\%m_{KClO_3} = \dfrac{0,2.122,5}{87,7}.100\% = 27,93\%$
$\%m_{KMnO_4} = 100\% -27,93\% = 72,07\%$
b) $m_{O_2} = 87,7 - 81,3 = 6,4(gam) \Rightarrow n_{O_2} = 0,2(mol)$
$n_{Cl_2} = 1,15(mol)$
Bảo toàn electron :
$6n_{KClO_3} + 5n_{KMnO_4} = 4n_{O_2} + 2n_{Cl_2}$
$\Rightarrow 6a + 5b = 0,2.4 + 1,15.2(3)$
Từ (1)(3) suy ra a = \(\dfrac{513}{3355}\) ; b = \(\dfrac{2929}{6710}\)
\(\%m_{KClO_3}=\dfrac{\dfrac{513}{3355}.122,5}{87,7}.100\%=21,36\%\)
$\%m_{KMnO_4} = 100\% - 21,36\% = 78,64\%$