Reakcja:
Na₂O + H₂O ⇒ 2 NaOH
m (masa) Na₂O- 124g
M (masa mola) Na₂O= 2*23g + 16g = 62g
M 2 moli NaOH = 2* (23g + 16g + 1g) = 80g
mw (mas wody) - 370g
Zgodnie z rakcją:
z 62g Na₂O-------uzyskamy-----80g NaOH
to ze 124g Na₂O--uzyskamy---- X g NaOH
X= 124g * 80g : 62g =160g NaOH(ms)
mr (masa roztworu)= m Na₂O + m H₂O = 124g + 370g = 494g
Obliczam stężenie procentowe (C%) roztworu NaOH:
C% = ms *100% : mr = 160g *100% : 494g ≈ 32,4 %
Na2O+H2O-->2NaOH62g-------------80g124g------------xx=160gms=160gmr=124+370=494gCp=ms*100%/mr=32,39%
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Reakcja:
Na₂O + H₂O ⇒ 2 NaOH
m (masa) Na₂O- 124g
M (masa mola) Na₂O= 2*23g + 16g = 62g
M 2 moli NaOH = 2* (23g + 16g + 1g) = 80g
mw (mas wody) - 370g
Zgodnie z rakcją:
z 62g Na₂O-------uzyskamy-----80g NaOH
to ze 124g Na₂O--uzyskamy---- X g NaOH
X= 124g * 80g : 62g =160g NaOH(ms)
mr (masa roztworu)= m Na₂O + m H₂O = 124g + 370g = 494g
Obliczam stężenie procentowe (C%) roztworu NaOH:
C% = ms *100% : mr = 160g *100% : 494g ≈ 32,4 %
Na2O+H2O-->2NaOH
62g-------------80g
124g------------x
x=160g
ms=160g
mr=124+370=494g
Cp=ms*100%/mr=32,39%