Răspuns:
Explicație:
Hristos a inviat!
Date
ms=200g NaOH, ω=20%
CO₂
Cerinte
m Na₂CO₃ =?g
n Na₂CO₃ =?moli
Rezolvare
ω/100= md/ms⇒md=ms×ω⇒
md=200×20/100⇒md=40g NaOH
2×40 106
2NaOH + CO₂ → Na₂CO₃ + H₂O
40 a
a=40×106/2×40=53 g Na₂CO₃
M NaOH = 23+16+1=40 g/mol
M Na₂CO₃ = 23×2+12+16×3=106 g/mol
1mol..........................106g Na₂CO₃
b.................................53g
b=53/106=1/2=0,5 moli Na₂CO₃
