Question

Liquid benzene burns in oxygen according to:
$2C_6{H}_6+15O_2\to 12C{O}_2+6H_{2}O$
If density of liquid benzene is 0.88 g/cc, what volume of $O_2$ at STP is needed to complete the combustion of 39 cc of liquid benzene ?

• A. 11.2 $litre$
• B. 74 $litre$
• C. 0.074 $m^3$
• D. 37 $d{m}^3$

Answer 1

Answer: (B)

74 $litre$

we have given the density of liquid benzene as $0.88g/cc$

that means, mass of 1 ml of benzene = 0.88 g

therefore mass of 39 ml of benzene = 0.88* 39 = 34.32 g

no. of moles of benzene = $\frac{givenweight}{molecularweight}$ = 0.44

acc to the chemical reaction

for 2 moles of benzene we require 15 mole of oxygen

therefore for 0.44 mole of benzene we require oxygen = $\frac{15}{2}\ast 0.44$ = 3.3

volume of 1 mole of gas at STP = 22.4 L

so, volume of 3.3 mole of oxygen = 22.4*3.3 = 73.92 L