дано
m( ppa Ba(NO3)2) = 2.61 g
W(Ba(NO3)2) = 10%
m(BaSO4) - ?
m(Ba(NO3)2) = 2.61 * 10% / 100% = 0.261 g
M(Ba(NO3)2) = 261 g/mol
n(Ba(NO3)2) = m/M = 0.261 / 261 = 0.001 mol
K2SO4+Ba(NO3)2 = 2KNO3+BaSO4↓
n(Ba(NO3)2) = n( BaSO4) = 0.001 mol
M( BaSO4) = 233 g/mol
m(BaSO4) = n*M = 0.001 * 233 = 0.233 g
ответ 0.233 г
Объяснение:
дано
m( ppa Ba(NO3)2) = 2.61 g
W(Ba(NO3)2) = 10%
m(BaSO4) - ?
m(Ba(NO3)2) = 2.61 * 10% / 100% = 0.261 g
M(Ba(NO3)2) = 261 g/mol
n(Ba(NO3)2) = m/M = 0.261 / 261 = 0.001 mol
K2SO4+Ba(NO3)2 = 2KNO3+BaSO4↓
n(Ba(NO3)2) = n( BaSO4) = 0.001 mol
M( BaSO4) = 233 g/mol
m(BaSO4) = n*M = 0.001 * 233 = 0.233 g
ответ 0.233 г
Объяснение: