Benzene burns in oxygen according to the equation 2C6H6(l) + 15O2(g) → 12CO2(g) + 6H2O(l). How many litres of oxygen are required at STP for the complete combustion of 39g of liquid benzene?

2C6H6(l) + 15O2(g) → 12CO2(g) + 6H2O(l). 2×78g 15×22.4L 39g ? Vol. of O2 required=392×78×15×22.4=84L

Benzene burns in oxygen according to the equation 2C6H6(l) + 15O2(g) → 12CO2(g) + 6H2O(l). How many litres of oxygen are required at STP for the complete combustion of 39g of liquid benzene?

2C6H6(l) + 15O2(g) → 12CO2(g) + 6H2O(l). 2×78g 15×22.4L 39g ? Vol. of O2 required=392×78×15×22.4=84L

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Benzene burns in oxygen according to the equation 2C₆H_6(l) + 15O_2(g) → 12CO_2(g) + 6H₂O_(l). How many litres of oxygen are required at STP for the complete combustion of 39g of liquid benzene?

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11.2

22.4

answer is D.

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$2 C_{6} H_{6} (l) + 15 O_{2} (g) \to 12 {CO}_{2} (g) + 6 H_{2} O (l) . 2 \times 78 g 15 \times 22.4 L 39 g ? Vol . of O_{2} required = \frac{39}{2 \times 78} \times 15 \times 22.4 = 84 L$

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Benzene burns in oxygen according to the equation 2C6H6(l) + 15O2(g) → 12CO2(g) + 6H2O(l). How many litres of oxygen are required at STP for the complete combustion of 39g of liquid benzene?

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Benzene burns in oxygen according to the equation 2C₆H_6(l) + 15O_2(g) → 12CO_2(g) + 6H₂O_(l). How many litres of oxygen are required at STP for the complete combustion of 39g of liquid benzene?