Theo đề bài ta có : nZn = 9,75/65 = 0,15(mol)
PTHH :
Zn + 2CH3COOH - > (CH3COO)2Zn + H2 \(\uparrow\)
0,15mol...0,3mol................0,15mol............0,15mol
a) mCH3COOH = 0,3.60 = 18(g)
=> mddCH3COOH = \(\dfrac{18.100}{60}=30\left(g\right)\)
b) C%dd(CH3COO)2Zn = \(\dfrac{0,15.183}{9,75+30-0,15.2}.100\) ≃ 69,58%