Odpowiedzi

2010-02-21T16:05:53+01:00
xg (nad Ch3COOH), 100g(nad NaOH)
CH3(indeks dolny)COOH+NaOH -> CH3(indeks dolny)COONa+H2(indeks dolny)O
1 mol (pod CH3COOH),1 mol(pod NaOH)
60g (pod CH3COOH), 40g(pod NaOH)

1 mol CH3COOH - 40 g NaOH
xg CH3COOH- 100 g NaOH

x= (100g*60g):40g= 150g