Question

Answer 1

a)

$Mg + 2CH_3COOH \to (CH_3COO)_2Mg + H_2$

$n_{(CH_3COO)_2Mg} = \dfrac{0,71}{142} = 0,005(mol)$

$n_{CH_3COOH} = 0,005.2 = 0,01(mol)$

$C_{M_{CH_3COOH}} = \dfrac{0,01}{0,025} = 0,4M$

b)

$CH_3COOH + NaOH \to CH_3COONa + H_2O$
$n_{NaOH}  = n_{CH_3COOH} = 0,01(mol)$
$m_{dd\ NaOH} = \dfrac{0,01.40}{10\%} = 4(gam)$