What volume of propane is burnt for every $500 c m^{3}$ of air used in the reaction under the same conditions?
(Assuming oxygen is $(\frac{1}{5})^{t h}$ of air)

$C_{3} H_{8} + 5 O_{2} \to 3 C O_{2} + 4 H_{2} O$

50 cm³

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100 cm³

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150 cm³

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175 cm³

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Solution

Step 1:
1 mole of propane combines with 5 moles of oxygen
$V o l u m e \propto N u m b e r o f m o l e s$
Step 2:
$1 {c m}^{3} o f p r o p a n e c o m b i n e s w i t h 5 {c m}^{3} o f O_{2}$
$500 {c m}^{3} o f a i r c o n t a i n s = \frac{500}{5} = 100 {c m}^{3} o f o x y g e n$
${5 c m}^{3} o f o x y g e n = {1 c m}^{3} o f p r o p a n e$
${100 c m}^{3} o f o x y g e n = \frac{100}{5} = {20 c m}^{3} o f p r o p a n e$

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What volume of Butane is burnt for every $1000 c m^{3}$ of air used in the reaction under the same conditions.

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