KOH + HCl = KCl + H₂O
[tex]md_{KOH} = \frac{ms*c}{100} = \frac{200*28}{100}= 56g [/tex] KOH
[tex]n = \frac{m}{M} = \frac{56}{56} = 1 mol [/tex] KOH => (cf. reactiei) 1mol HCl => ( cf. reactiei ) 1 mol KCl
[tex]n = \frac{m}{M} <=> m = n*M = 1*36.5 = 36.5g HCl [/tex] = md
[tex] ms_{HCl} = \frac{md*100}{c}= \frac{36.5*100}{30}= 121,66 g (solutie)[/tex]
[tex]m_{KCL} = n*M = 1*74,5 = 74,5g =md [/tex]
[tex]ms_{f} = 200 + 121,66 = 321,66g [/tex]
[tex]c_{f} = \frac{md*100}{ms_{f}} = \frac{74,5*100}{321,66}[/tex] = 23,16%
