Question

Answer 1

a)

Coi $n_{H_2O} = 3(mol) \Rightarrow n_H = 3.n_{H_2O} = 9(mol)$

Bảo toàn H :

$n_{H_3PO_4} = \dfrac{9-3.2}{3} = 1(mol)$
Ta có :
$C\%_{H_3PO_4} = \dfrac{1.98}{1.98 + 3.18}.100\% = 64,47\%$

b)

$n_P = n_{H_3PO_4} = \dfrac{200.64,47\%}{98} = 1,32(mol)$

$m_P = 1,32.31 = 40,92(gam)$

c)

$n_{H_3PO_4} = \dfrac{1,32}{2} = 0,66(mol)$

$2NaOH + H_3PO_4 \to Na_2HPO_4 + 2H_2O$

$n_{NaOH} = 2n_{H_3PO_4} = 1,32(mol)$
$V_{dd\ NaOH} = \dfrac{1,32.40 : 20\%}{1,225} = 215,51(ml)$