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Chemistry Question on Organic Chemistry- Some Basic Principles and Techniques

Liquid benzene (C6H6)(C_6H_6) bums in oxygen according to the equation,       2C6H6(I)+15O2(g)→12CO2(g)+6H2O(g)\, \, \, \, \, \, 2C_6H_6 (I) + 15O_2 (g) \rightarrow 12 CO_2 (g) + 6H_2O(g) How many litres of O2O_2 at STP are needed to complete the combustion of 39 g of liquid benzene? (Mol. weight of O2=32,C6H6=75)(Mol. \, weight \, of \, O_2 = 32, C_6H_6 = 75)

A

74 L

B

11.2 L

C

22.4 L

D

84 L

Answer

84 L

Explanation

Solution

(d) 2C6H6+15O2(g)→12CO2(g)+6H2O(g)2C_6 H_6 + 15O _2 (g) \rightarrow 12CO_2 (g) + 6H_2O(g) =1562×78     =33015×32_{=156}^{2 \times 78} \, \, \, \, \, _{=330}^{15 \times 32} ∴\therefore 156 g of benzene required oxygen =15×22.4L= 15 \times 22.4 L ∴\therefore 1 g of benzene required oxygen =15×22.4156L= \frac {15 \times 22.4} {156}L ∴\therefore 39 g of benzene required oxygen =15×22.4×39156 = \frac{15 \times 22.4 \times 39} {156} = 84.0 L