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. (a)`81 g`
B. (b)`40.5 g`
C. (c )`20.25 g`
D. (d)`162 g`

Answer

B

Explanation

`Ba(OH)_(2)+CO_(2) rarr Ba CO_(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)`
`:. `205 moles 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 g~~ 40.5 g`