A) 112g
B) 84g
C) 56g
D) 28g
Correct Answer: B
Solution :
[b] \[CO+\frac{1}{2}{{O}_{2}}\to C{{O}_{2}};\]\[C{{O}_{2}}+2KOH\to {{K}_{2}}C{{O}_{3}}+{{H}_{2}}O\] |
Moles of \[KOH=\frac{28}{56}=0.50\] |
It corresponds to 0.25 mol of \[C{{O}_{2}}\] |
Hence mol of \[CO=1-0.25=0.75\]\[\equiv \] mole of \[C{{O}_{2}}\]formed |
Mol of KOH requred \[=2\times 0.75=1.5\]\[=1.5\times 56=84g\] |
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