Ответ:
дано
m( ppaBa(NO3)2) = 26.1 g
W(Ba(NO3)2) = 10%
----------------------
m(BaSO4) - ?
m(Ba(NO3)2) = 26.1*10% / 100% = 2.61 g
Ba(NO3)2+Na2SO4-->BaSO4↓+2NaNO3
M(Ba(NO3)2) = 261 g/mol
n(Ba(NO3)2) = m/M = 2.61 / 261 = 0.01 mol
n(Ba(NO3)2) = n(BaSO4) = 0.01 mol
M(BaSO4) = 233 g/mol
m(BaSO4) = n*M = 0.01 * 233 = 2.33 g
ответ 2.33 гр.
Объяснение:
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Answers & Comments
Ответ:
дано
m( ppaBa(NO3)2) = 26.1 g
W(Ba(NO3)2) = 10%
----------------------
m(BaSO4) - ?
m(Ba(NO3)2) = 26.1*10% / 100% = 2.61 g
Ba(NO3)2+Na2SO4-->BaSO4↓+2NaNO3
M(Ba(NO3)2) = 261 g/mol
n(Ba(NO3)2) = m/M = 2.61 / 261 = 0.01 mol
n(Ba(NO3)2) = n(BaSO4) = 0.01 mol
M(BaSO4) = 233 g/mol
m(BaSO4) = n*M = 0.01 * 233 = 2.33 g
ответ 2.33 гр.
Объяснение: