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6 g ---------- x g
60 g --------- 40 g
x = 4 g
mCH3COOH = 60g
mNaOH = 40g
W jednym molu mamy 60g CH3COOH i 40g NaOH, tak więc układamy proporcję:
60 g CH3COOH ----- 6g
40g NaOH ------- xg
60x = 40g * 6g
60x = 240g / : 60
x = 4g
Potrzeba 4g NaOH
60g ------ 40g
6g ------ xg
x = 4 g
Odp. Potrzeba 4g NaOH.