Question

Answer 1

a) 

$ZnO + 2CH_3COOH \to (CH_3COO)_2Zn + H_2O$
$Fe + 2CH_3COOH \to (CH_3COO)_2Fe + H_2$

Theo PTHH : 

$n_{Fe} = n_{H_2} = \dfrac{2,479}{24,79} = 0,1(mol)$

$\%m_{Fe} = \dfrac{0,1.56}{13,7}.100\% = 40,9\%$

$\%m_{ZnO} = 100\% - 40,9\% = 59,1\%$

b) $n_{ZnO} = \dfrac{13,7 - 0,1.56}{81} = 0,1(mol)$
$n_{(CH_3COO)_2Zn} = n_{ZnO} = 0,1(mol)$

$n_{(CH_3COO)_2Fe} = n_{Fe} = 0,1(mol)$
$C_{M_{(CH_3COO)_2Fe}} = \dfrac{0,1}{0,2}  = 0,5M$

$C_{M_{(CH_3COO)_2Zn}} = \dfrac{0,1}{0,2}  = 0,5M$