Liquid benzene (C₆H₆) burns in oxygen according to the equation 2 C₆H₆(l) + 15 O₂(g) → 12 CO₂(g) + 6 H₂O(g) How many litres of O₂ at STP are needed to complete the combustion of 39 g of liquid benzene? (Mol. wt. of O₂ = 32, C₆H₆ = 78)

Question

Accepted Answer

112 L. The molar mass of benzene (C₆H₆) is 78 g/mol, so 39 g corresponds to 0.5 moles. From the balanced equation, 2 moles of C₆H₆ require 15 moles of O₂. Thus, 0.5 moles of C₆H₆ will require (15/2) × 0.5 = 3.75 moles of O₂. At STP, 1 mole of gas occupies 22.4 L, so 3.75 moles of O₂ will occupy 3.75 × 22.4 = 84 L. Hence, the correct answer is 84 L.

Liquid benzene (C₆H₆) burns in oxygen according to the equation 2 C₆H₆(l) + 15 O₂(g) → 12 CO₂(g) + 6 H₂O(g) How many litres of O₂ at STP are needed to complete the combustion of 39 g of liquid benzene? (Mol. wt. of O₂ = 32, C₆H₆ = 78)

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