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5 moles of Ba(OH)(2) are treated with ex...

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

A

39.4 g

B

197 g

C

591 g

D

985g

Text Solution

Verified by Experts

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
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