ЛС
Липницкая Светлана
Ba(OH)2+H2SO4 ---> BaSO4 + 2H2O
M(Ba(OH)2) = 137 + 16*2 + 2*2 =173 g/mol
m (Ba(OH)2) = 24 gramm (uslovie)
M(BaSO4) = 137+ 32 + 16*4 = 233 g/mol
m (Ba(OH)2) / M(Ba(OH)2) = m (BaSO4) / M(BaSO4)
m (BaSO4) = 233*24/173 = 32.32 g