The volume of oxygen at STP required to burn 2.4 g of carbon completely is
(a) 1.12 L
(b) 8.96 L
(c) 2.24 L
(d) 4.48 L
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Answer is : (d) 4.48 L
C(s) + O2(g) → CO2(g)
1 mole of carbon reacts completely with 1 mole of oxygen to produce 1 mole of CO2.
12 g of C reacts = 1 mole of O2
2.4 g of C reacts with = 1/12 x 2.4 = 0.2 mole of O2
At STP 1 mole of O2 contains = 22.4 L
∴ 0.2 mole of O2 contains = 22.4 x 0.2 = 4.48 L