Calculate the mass of `BaCO_(3)` produced when excess `CO_(2)` is bubbled through a solution containing `0.205` moles of `Ba(OH)_(2)`.

Options

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

Answer

B

Explanation

`Ba(OH_(2)) + CO_(2) to BaCO_(3) +H_(2)O`
Atomic wt. of `BaCO_(3) = 137+12+16 xx 3 = 197`
No.of mole ` = ("wt.of substance")/("mol wt.")`
`because` 1 mole of `Ba(OH)_(2)` gives 1 mole of `BaCO_(3)`
`therefore` 0.205 mole of `Ba(OH)_(2)` will give 0.205 mole of `BaCO_(3)`
`therefore ` wt.of 0.205 mole of `BaCO_(3)` will be
` 0.205 xx 197 = 40.358 gm ~~ 40.5 g `