A) 81 g
B) 40.5 g
C) 20.25 g
D) 162 g
Correct Answer: B
Solution :
[b] \[\underset{1\,mol}{\mathop{Ba{{\left( OH \right)}_{2}}}}\,+C{{O}_{2}}\xrightarrow{{}}\underset{1\,mol}{\mathop{BaC{{O}_{3}}}}\,+{{H}_{2}}O\] |
\[1\,mol\,Ba{{(OH)}_{2}}=1mol\,BaC{{O}_{3}}\] |
\[\therefore 0.205\text{ }mol\text{ }Ba{{\left( OH \right)}_{2}}=0.205\text{ }mol\text{ }BaC{{O}_{3}}\] |
Wt. of substance = No. of moles x Molecular |
Mass \[=0.205\times 197.3=40.5\text{ }g\] |
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