Question

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Answer 1

a)

$n_{NaHCO_3} = \dfrac{42}{84} = 0,5(mol)$

$NaHCO_3 + CH_3COOH \to CH_3COONa + CO_2 + H_2O$

Theo PTHH : 

$n_{CH_3COOH} = n_{NaHCO_3} = 0,5(mol)$
$m_{dd\ CH_3COOH} = \dfrac{0,5.60}{12\%} = 250(gam)$

b) $n_{CO_2} = n_{NaHCO_3} = 0,5(mol)$
$V_{CO_2} = 0,5.22,4 = 11,2(lít)$

c) $CH_3COOH + NaOH \to CH_3COONa + H_2O$
$n_{NaOH} = n_{CH_3COOH} = 0,5(mol)$
$V_{dd\ NaOH} = \dfrac{0,5}{0,2} = 2,5(lít)$