дано
m(ppa Ca(OH)2) = 1.94 g
W(Ca(OH)2)= 20%
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m(CaSO4)-?
m(Ca(OH)2) = 1.94 * 20% / 100% = 0.388 g
M(Ca(OH)2) = 74 g/mol
Ca(OH)2+H2SO4-->CaSO4+2H2O
n(Ca(OH)2) = m/M = 0.388 / 74 = 0.005 mol
n(Ca(OH)2) = n(CaSO4) = 0.005 mol
M(CaSO4) = 136 g/mol
m(CaSO4) = n*M = 0.005 * 136 = 0.68 g
ответ 0.68 г