Помогите с задачей К раствору серной кислоты массой 150 гр с

дано

m(ppa H2SO4) = 150 g

W(H2SO4) = 20%

+KOH

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

m(K2SO4)-?

m(H2O)-?

m(H2SO4) = m(ppa H2SO4) * W(H2SO4) / 100%

m(H2SO4) = 150 * 20% / 100% = 30 g

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

M(H2SO4)  = 98 g/mol

n(H2SO4) = m(H2SO4) / M(H2SO4) = 30 / 98 = 0.31 mol

n(H2SO4) = n(K2SO4) = 0.31 mol

M(K2SO4) = 174 g/mol

m(K2SO4) = n(K2SO4) * M(K2SO4) = 0.31 * 174 = 53.94 g

n(H2SO4) = 2n(H2O)

n(H2O) = 2* 0.31 = 0.62 mol

M(H2O) = 18 g/mol

m(H2O) = n(H2O) * M(H2O) = 0.62 * 18 = 11.16 g

ответ 53.94 г , 11.16 г