m(р-ну) = 285 g
W(Ba(OH)2) = 12%
m(BaCO3 практ.) = 37.43 g
η(BaCO3) - ?
m(Ba(OH)2) = 0.12 * 285 = 34.2 g
n(Ba(OH)2) = 34.2 / 171 = 0.2 mol
0.2 mol x mol
Ba(OH)2 + CO2 = BaCO3 + H2O
1 mol 1 mol
1:0.2 = 1:x
x = 0.2
m(BaCO3 теоритич.) = 0.2 * 197 = 39.4 g
η(BaCO3) = 37.43 / 39.4 = 0.95 = 95%
ответ: 95% .
m(р-ну) = 285 g
W(Ba(OH)2) = 12%
m(BaCO3 практ.) = 37.43 g
η(BaCO3) - ?
m(Ba(OH)2) = 0.12 * 285 = 34.2 g
n(Ba(OH)2) = 34.2 / 171 = 0.2 mol
0.2 mol x mol
Ba(OH)2 + CO2 = BaCO3 + H2O
1 mol 1 mol
1:0.2 = 1:x
x = 0.2
m(BaCO3 теоритич.) = 0.2 * 197 = 39.4 g
η(BaCO3) = 37.43 / 39.4 = 0.95 = 95%
ответ: 95% .