1.

The mass of BaCO_(3) produced when excess CO_(2) is bubbled through a solution of 0.205 mol Ba(OH)_(2) is

Answer»

81 g
40.5 g
20.25 g
162 g

Solution :`BA(OH)_(2)+CO_(2) rarr BaCO_(3)+H_(2)O`
Atomic wt. of `BaCO_(3)=137+12+16xx3=197`
No. of mole `=("wt. of substance")/("mol wt.")`
`:'` 1 mole of `Ba(OH)_(2)` GIVES 1 mole of `BaCO_(3)`
`:.` 0.205 mole of `Ba(OH)_(2)` will give 0.205 mole of `BaCO_(3)`
`:.` wt. of 0.205 mole of `BaCO_(3)` will be
`0.205xx197=40.385 GM ~~ 40.5 gm`.


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