A. (3a + 2b) mol;

B. (3,2a + 1,6b) mol;

C. (1,2a + 3b) mol;

D. (4a + 3,2b) mol.

Đáp án đúng là: D

P + 5HNO₃ → H₃PO₄ + 5NO₂↑ + H₂O

a….5a…………a                                 (mol)

S + 6HNO₃ → H₂SO₄ + 6NO₂↑ + 2H₂O

b…..6b……….b                                     (mol)

$\sum n_{HN{O}_{3}pu}=5a+6b\left(mol\right)$

$n_{HN{O}_3}$_dư = (5a + 6b). 20% = a + 1,2b (mol)

$\sum n_{H^{+}}=n_{HN{O}_3}+2n_{H_{2}S{O}_4}+3n_{H_{3}P{O}_4}$ = (a + 1,2b) + 2b + 3a = 4a + 3,2b (mol)

Phản ứng trung hoà: H⁺  + OH^- → H₂O

$\Rightarrow n_{NaOH}=n_{O{H}^{-}}=n_{H^{+}}=4a+3,2b$  (mol).