Question

Answer 1

Câu 1 : 

$n_{CaCO_3} = \dfrac{20}{100} = 0,2(mol)$
$n_{HCl} = \dfrac{91,25.20\%}{36,5} = 0,5(mol)$
             CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O

Ban đầu : 0,2.........0,5

Pứ : .........0,2.........0,4......................0,2...........................

Dư : ..........0...........0,1

$V_{CO_2} = 0,2.22,4 = 4,48(lít)$
$ m_{dd} = 200 + 91,25 - 0,2.44 = 102,45(gam)$
$C\%_{CaCl_2} = \dfrac{0,2.111}{102,45}.100\% = 21,67\%$

$C\%_{HCl} = \dfrac{0,1.36,5}{102,45}.100\% = 3,56\%$

Answer 2

Câu 2 : 

$n_{Fe_3O_4} = \dfrac{23,2}{232} = 0,1(mol)$
$n_{H_2SO_4} = \dfrac{245.20\%}{98}= 0,5(mol)$
              Fe₃O₄  + 4H₂SO₄ → Fe₂(SO₄)₃ + FeSO₄ + 4H₂O

Ban đầu: 0,1.............0,5...................................................

Phản ứng:0,1............0,4.............................................

Sau pư :   0................0,1..............0,1................0,1...............

$m_{dd} = 23,2 + 245 = 268,2(gam)$
$C\%_{H_2SO_4\ dư} = \dfrac{0,1.98}{268,2}.100\% = 3,65\%$
$C\%_{FeSO_4} = \dfrac{0,1.152}{268,2}.100\% = 5,67\%$

$C\%_{Fe_2(SO_4)_3} = \dfrac{0,1.400}{268,2}.100\% = 14,91\%$