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)>n(H2SO4)
n(H2SO4) = n(Na2SO4)
2NaOH+H2SO4-->Na2SO4+2H2O
M(Na2SO4) = 142 g/mol
m(Na2SO4) = n*M = 1*142 = 142 g
ответ 142 г