$n KMnO4 = 0,125 mol $$3C2H4 + 2KMnO4 + 4H2O =>
3C2H4(OH)2 + 2MnO2 + 2KOH$ $\frac{0,125 \times 80}{100}$ => $0,15$$=> m C_{2}H_{4}(OH)_{2} = 9,3 gam$
$n KMnO4 = 0,125 mol $
$3C2H4 + 2KMnO4 + 4H2O =>
3C2H4(OH)2 + 2MnO2 +
2KOH$
$\frac{0,125 \times 80}{100}$
=>
$0,15$
$=> m C_{2}H_{4}(OH)_{2} = 9,3 gam$