Chemistry · Chemical equations and stoichiometry
Liquid benzene burns in oxygen according to: If density of liquid benzene is wha
Liquid benzene burns in oxygen according to: \( 2 C_{6} H_{6}+15 O_{2} \rightarrow 12 C O_{2}+6 H_{2} O \) If density of liquid benzene is \( 0.88 \mathrm{g} / \mathrm{cc} \) what volume of \( \boldsymbol{O}_{2} \) at \( \mathrm{STP} \) is needed to complete the combustion of 39 cc of liquid benzene?
- A. 11.2 litre
- B. 74 litre
- C. \( 0.074 \mathrm{m}^{3} \)
- D. 37 \( d m^{3} \)
Step-by-step solution
Mass of benzene = density × volume = 0.88 g/cc × 39 cc = 34.32 g. Moles of benzene = 34.32 g / 78 g/mol = 0.44 mol. From the balanced equation, 2 mol C6H6 require 15 mol O2, so moles of O2 = 0.44 × (15/2) = 3.3 mol. At STP, volume of 1 mol gas = 22.4 L, so volume of O2 = 3.3 × 22.4 = 73.92 L ≈ 74 L.
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