Люди, упрашиваю! Помогите пожалуйста решить задачку!

дано

Vppa(NaOH) = 0.5L = 500 mL

W(NaOH) = 20 %

p(NaOH) = 1.22 g/sm^3

m(H2SO4) = 0.5 kg = 500 g

W(H2SO4) = 19.6 %

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m(Na2SO4)-?

m(NaOH) = W(NaOH)*Vppa(NaOH) * p(NaOH) / 100% = 20% * 500 * 1.22 / 100%

m(NaOH) = 122 g

m(H2SO4) = mppa(H2SO4) *W(H2SO4) / 100% = 19.6*500 / 100% = 98 g

M(NaOH) = 40 g/mol

n(NaOH) = m/M = 122 / 40 = 3.05 mol

M(H2SO4) = 98 g/mol

n(H2SO4) = m/M = 98 /98 = 1 mol

n(NaOH)gt;n(H2SO4)

n(H2SO4) = n(Na2SO4)

2NaOH+H2SO4--gt;Na2SO4+2H2O

M(Na2SO4) = 142 g/mol

m(Na2SO4) = n*M = 1*142 = 142 g

ответ 142 г