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

Options

A. 81 g
B. 40.5 g
C. 20.25 g
D. 162 g

Answer

B

Explanation

`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`.