Question

Determinati partea de masa a sulfatului de sodiu obtinut la amestecarea solutiei de hidroxid de sodium de 8%, v=50 ml si densitatea de1,1 g/cm(cub)
cu acid sulfuric cu 27%, v=20 ml si densitatea 1,2 g/cm(cub)

Answer 1

C₁= 8% NaOH
V₁ = 50mL
ρ₁ = 1,1g/cm³
C₂ = 27% H₂SO₄
V₂ = 20mL
ρ₂ = 1,2g/cm³

ρ₁ = ms₁/V₁ ⇒ ms₁ = 1,1×50 = 55g sol. NaOH
C₁ = md₁/ms₁×100 ⇒ md₁ = 8×55/100 = 4,4g

m H₂O = ms-md = 55-4,4 = 50,6g

ρ₂ = ms₂/V₂ ⇒ ms₂ = 1,2×20 = 24g sol. H₂SO₄

C₂ = md₂/ms₂×100 ⇒ md₂ = 27×24/100 = 6,48g H₂SO₄

m H₂O = ms-md = 24-6,48 = 17,52g

n NaOH = 4,4/40 = 0,11 moli
n H₂SO₄ = 6,48/98 = 0,066 moli 

2NaOH + H₂SO₄ ⇒ Na₂SO₄ + 2H₂O

2 moli NaOH..............1 mol H₂SO₄
0,11 moli NaOH.........x moli H₂SO₄

x = 0,055 moli H₂SO₄

2 moli NaOH.........142g Na2SO4
0,11 moli NaOH.......y g Na2SO4

y = 142×0,11/2 = 7,81g Na2SO4

n acid exces = 0,066-0,055 = 0,0111 moli 

m acid exces = 0,0111×98 = 1,09g 

mdf = 7,81g + m acid exces = 7,81+1,09 = 8,9g

msf = 8,9+0,11×18 + 50,6 + 17,52 = 79g

Cf = mdf/msf×100 = 8,9×100/79 ≈ 11,265%