A) 81 g
B) 40.5 g
C) 20.25 g
D) 162 g
Correct Answer: B
Solution :
\[Ba{{(OH)}_{2}}+C{{O}_{2}}\to BaC{{O}_{3}}+{{H}_{2}}O\] Atomic wt. of \[BaC{{O}_{3}}\] =\[137+12+16\times 3\]= 197 No. of mole \[=\frac{\text{wt}\text{. of substance}}{\text{mol wt}\text{.}}\] \[\because \] 1 mole of \[Ba{{(OH)}_{2}}\] gives 1 mole of \[BaC{{O}_{3}}\] \[\therefore \] 205 mole of \[Ba{{(OH)}_{2}}\]will give .205 mole of \[BaC{{O}_{3}}\] \[\therefore \] wt. of 0.205 mole of \[BaC{{O}_{3}}\] will be \[.205\times 197=40.385gm\approx \,40.5\,gm\]
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