Задание 1. Возьмите химические стаканы и в каждый налейте воду и напитки для анализа. Погрузите универсальный индикатор в стаканы. Представь- те полученные результаты в таблице.

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S + O2 = SO2
2SO2 + O2 = 2SO3
SO3 + H2SO4 = H2S2O7 (олеум)
m(H2SO4 р-р) = V(H2SO4)xp(H2SO4) = 400x1,14 = 456 г, m(H2SO4) = m(H2SO4 р-р)xW(H2SO4) = 456x0,2 = 91,2 г, m(H2S2O7 р-р) = m(H2SO4 р-р)xW(H2SO4)/W(H2S2O7) = 456x20/60 = 152 г. Т.к. 60% раствор получился добавлением SO3, то m(SO3) = m(H2S2O7 р-р) - m(H2SO4) = 152 - 91,2 = 60,8 г. По уравнениям реакции видно, что n(S) = n(SO2) = n(SO3) = m(SO3)/M(SO3) = 60,8/80 = 0,76 моль, где M(SO3) = 32+3x16 = 80 г/моль. m(S) = n(S)xM(S) = 0,76x32 = 24,32 г.