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InterviewSolution
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A mixture of 2.3 g formic acid and 4.5 g oxalic acid is treated with conc. `H_(2)SO_(4)`. The evolved gaseous mixture is passed through KOH pellets. Weight (in g) of the remaining product at STP will beA. 2.8B. `3.0`C. 1.4D. 4.4 |
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Answer» Correct Answer - A Key Concept firstly, write the reaction of formic acid and oxalic acid with conc. `H_(2)SO_(4)`, respectively. Then, find the gaseous products formed and identify the remaining gaseous product after passing through KOH. Finally, calculate the total number of moles of gaseous product. `underset("Formic acid")(HCO OH)overset(Conc. H_(2)SO_(4))to CO(g)+H_(2)O(l)` `{:("Initial moles",(2.3)/(46)=(1)/(20)mol,0),("Final moles"," "0,(1)/(20)):}` Similarly, `{:(CO OH),(|),(CO OH),("Oxalic acid"):}overset(Conc.H_(2)SO_(4))toCO(g)+CO_(2)(g)+H_(2)O(l)` `{:("Initial moles",(4.5)/(90)=(1)/(20)mol,0,0),("Final moles"," "0,(1)/(20),(1)/(20)):}` Now, `H_(2)O(l)` gets absorbed by conc. `H_(2)SO_(4)`. gaseous mixture CO and `CO_(2)` when passed through KOH, only `CO_(2)` gets absorbed. Thus, CO is the remaining gas. Total number of moles of CO formed in the above equation `=(1)/(20)+(1)/(20)=(1)/(10)` `because " " " Moles"=("Weight")/("Molar mass")` `therefore " Weight of CO formed" =(1)/(10)xx25=2.8 g` Thus, weight of the remaining product at STP will be 2.8 g. |
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