Question

200 g sol KOH de c=28% reactioneaza cu o sol de HCl de c=30% . Cati moli de acid s-au consumat , cati moli de sare rezulta si ce concentratie are solutia sarii rezultate daca reactia a fost totala ?

Answer 1

KOH + HCl  =  KCl + H₂O

$md_{KOH} = \frac{ms*c}{100} = \frac{200*28}{100}= 56g$ KOH

$n = \frac{m}{M} = \frac{56}{56} = 1 mol$ KOH => (cf. reactiei) 1mol HCl => ( cf. reactiei ) 1 mol KCl

$n = \frac{m}{M} <=> m = n*M = 1*36.5 = 36.5g HCl$ = md

$ms_{HCl} = \frac{md*100}{c}= \frac{36.5*100}{30}= 121,66 g (solutie)$

$m_{KCL} = n*M = 1*74,5 = 74,5g =md$

$ms_{f} = 200 + 121,66 = 321,66g$

$c_{f} = \frac{md*100}{ms_{f}} = \frac{74,5*100}{321,66}$ = 23,16%