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Question

12.6 g nonene C9H18 is combusted in the presence of air. The percentage of oxygen in air is 21 percent by volume at STP.  The volume of air needed for complete combustion of nonene is


Your Answer
A
112 L 
Correct Answer
B
144 L 
Your Answer
C
22.4 L 
Your Answer
D
224 L

Solution

The correct option is B 144 L 
Balanced equation is
C9H18+272O29CO2+9H2O
Given: Mass of nonene = 12.6 g
Number of moles of nonene =12.6126=0.1 mol 
Moles of O2 at  STP =2721×0.1
                                    =2.72 mol
Volume of O2 at STP =2.72×22.4=2.7×11.2 L
21 L of oxygen is present 100 L of air
volume of air required =10021×2.7×11.2=144 L

 

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