Question

Answer 1

H2SO4+2NaOH->Na2SO4+H2O

0,02-----0,04--------0,02 mol

n H2SO4=1.0,02=0,02 mol

=>m NaOH=0,04.40=1,6 g

=>m dd NaOH=8g

m Na2SO4=0,02.142=2,84 g

Answer 2

\(n_{H_2SO_4}=0,02.1=0,02\left(mol\right)\)

PTHH: 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

Mol:      0,04            0,02           0,02

\(m_{ddNaOH}=\dfrac{0,04.40.100}{20}=8\left(g\right)\)

\(m_{Na_2SO_4}=0,02.142=2,64\left(g\right)\)