Some Basic Concepts Of Chemistry Question 66
Question: 1 mole of mixture of CO and $ CO _2 $ requires exactly $ 28gKOH $ in solution for complete conversion of all the $ CO _2 $ into $ K _2CO _3. $ How much amount more of KOH will be required for conversion into $ K _2CO _3. $ If one mole of mixture is completely oxidized to $ CO _2 $
Options:
A) 112g
B) 84g
C) 56g
D) 28g
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ CO+\frac{1}{2}O _2\to CO _2; $ $ CO _2+2KOH\to K _2CO _3+H _2O $ Moles of $ KOH=\frac{28}{56}=0.50 $ It corresponds to 0.25 mol of $ CO _2 $ Hence mol of $ CO=1-0.25=0.75 $ $ \equiv $ mole of $ CO _2 $ formed Mol of KOH requred $ =2\times 0.75=1.5 $ $ =1.5\times 56=84g $
