8g tlenku miedzi (II) przereagowało całkowicie z kwasem octowym. Oblicz, ile octanu miedzi otrzymano.
2CH3COOH + CuO ----> (CH3COO)2Cu + H2O
80g CuO ---- 182g (CH3COO)2Cu
8g CuO ---- X
X=18,2g (CH3COO)2Cu
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2CH3COOH + CuO ----> (CH3COO)2Cu + H2O
80g CuO ---- 182g (CH3COO)2Cu
8g CuO ---- X
X=18,2g (CH3COO)2Cu