Question

Hòa tan a gam Na2O vào m gam H2O thu được 400 gam dd NaOH 1,15%.Tính a,m

Answer 1

\(n_{Na_2O}=\dfrac{a}{62}\left(mol\right)\)

\(m_{NaOH}=\dfrac{400.1,15}{100}=4,6\left(g\right)\)

PTHH: Na₂O + H₂O --> 2NaOH

              \(\dfrac{a}{62}\)------------->\(\dfrac{a}{31}\)

=> \(m_{NaOH}=\dfrac{40a}{31}=4,6\left(g\right)\) => a = 3,565 (g)

=> m_{dd sau pư} = a + m = 400 (g) => m = 396,435 (g)