A) 4 : 1
B) 1 : 4
C) 1 : 2
D) 2 : 1
Correct Answer: A
Solution :
Suppose mole of\[HCOOH=x\] Mole of \[{{H}_{2}}{{C}_{2}}{{O}_{4}}=y\] \[\underset{x}{\mathop{HCOOH}}\,+{{H}_{2}}S{{O}_{4}}\to \underset{x}{\mathop{CO}}\,+{{H}_{2}}O+{{H}_{2}}S{{O}_{4}}\] \[\underset{y}{\mathop{{{H}_{2}}{{C}_{2}}{{O}_{4}}}}\,+{{H}_{2}}S{{O}_{4}}\to \underset{y}{\mathop{CO}}\,+\underset{y}{\mathop{C{{O}_{2}}}}\,+{{H}_{2}}O+{{H}_{2}}S{{O}_{4}}\] Total gaseous products = x + 2y Only \[C{{O}_{2}}\] is absorbed by KOH, y mole Hence, \[\frac{y}{x+2y}=\frac{1}{6}\] \[\therefore \]\[x+2y=6y\] \[x=4y\] \[\therefore \]\[\frac{x}{y}=\frac{4}{1}\]
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