5 moles of Ba(OH)_(2) are treated with excess of CO_(2). How much Ba(OH_(2) will be formed?

5 moles of Ba(OH)_(2) are treated with excess of CO_(2). How much Ba(O

1.	5 moles of Ba(OH)_(2) are treated with excess of CO_(2). How much Ba(OH_(2) will be formed?
Answer» 39.4 g 197 g 591 g 985g Solution :d) `BA(OH)_(2) + CO_(2) to BaCO_(3) + H_(2)O` `therefore` 5 moles of `Ba(OH)_(3)` = 5 moles of `BaCO_(3)` `therefore` MASS of `BaCO_(3)` = Moles of BaCO_(3) XX Molecular mass of `BaCO_(3)` = `5 xx 197) = 985 g