При реакции нейтрализации 200г 9,8%-ного раствора серной кислоты и 400г 7%-ного

дано

m(ppa H2SO4) =200 g

W(H2SO4) = 9.8 %

m(ppa KOH) = 400 g

w(KOH) = 7%

--------------------------

m(K2SO4)-?

m(H2SO4) = 200 * 9.8% / 100% = 19.6 g

M(H2SO4) = 98 g/mol

n(H2SO4) = m/M = 19.6 / 98 = 0.2 mol

m(KOH) = 400*7% / 100% = 28 g

M(KOH) = 56 g/mol

n(KOH) = m/M = 28 / 56 = 0.5 mol

n(H2SO4) lt; n(KOH)

H2SO4+2KOH--gt;K2SO4+2H2O

n(H2SO4) = n(K2SO4) = 0.2 mol

M(K2SO4) = 174 g/mol

m(K2SO4) = n*M = 0.2*174 = 34.8 g

ответ 34.8 г