Question

Cati litri de solutie de 0.5 N de k2cr2o7 sunt necesari oxidarii a 132 g de 3hidroxibutanona?

Answer 1

CH3-CO-CH(OH)-CH3 + [O] ⇒ CH3-CO-CO-CH3 + H2O 
K2Cr2O7 + 4H2SO4 ⇒ K2SO4 + Cr2(SO4)3 + 4H2O + 3[O]
2Cr⁶⁺ + 6e⁻ = 2Cr³⁺    E K2Cr2O7 = M/6     e = 6n
n cetona/nK2Cr2O7 = 3/1
(132/88)/nK2Cr2O7  = 3/1   nK2Cr2O7 = 0,5 moli;  eK2Cr2O7 = 3 vali 
Vs = e/cN = 3/0,5 = 6 l