A) 7 L
B) 6 L
C) 5 L
D) 10 L
Correct Answer: C
Solution :
Key Idea : Volume of a gas at STP = 22.4 L \[{{C}_{3}}{{H}_{8}}+5{{O}_{2}}\xrightarrow[{}]{{}}3C{{O}_{2}}+4{{H}_{2}}O\] \[22.4L\]\[5\times 22.4L\] \[\because \]Tb burn 22.4 L \[{{C}_{3}}{{H}_{8}}\] the oxygen required is \[=5\times 22.4L\] \[\therefore \]To bum \[1L{{C}_{3}}{{H}_{8}}\] the oxygen required will be \[=\frac{5\times 22.4}{22.4}\]\[=5L\]
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