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Calculate the volume of oxygen required for the complete combustion of 8.8 g of propane (C3H5). (Atomic mass: C = 14, O = 16, H = 1, Molar Volume = 22.4 dm3 at STP.)

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Question

Calculate the volume of oxygen required for the complete combustion of 8.8 g of propane (C3H5).
(Atomic mass: C = 14, O = 16, H = 1, Molar Volume = 22.4 dm3 at STP.)

Numerical
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Solution

\[\ce{C3H8 + 5O2 -> 3CO2 + 4H2O}\]

Molar weight of propane = 12 × 3 + 8 = 44

Volume of 5O2 = 5 × 22.4 = 112 litres

44 g of propane requires = 112 litres of oxygen

1 g of propane requires = `112/44` litres

8.8 g of propane requires = `112/44 xx 8.8`

= 112 × 0.2

= 22.4 litres

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Chapter 5: Mole Concept and Stoichiometry - Exercise 10 [Page 124]

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Frank Chemistry Part 2 [English] Class 10 ICSE
Chapter 5 Mole Concept and Stoichiometry
Exercise 10 | Q 1.3 | Page 124

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