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

= 84 litVolume of O2 required for the combustion of 39grams of C6H6 = 84 lit

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

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Detailed Solution

$large mathop {mathop {2{C_6}{H_6}(l)}limits_{(2 times 78g)}^{2moles} }limits_{39g} + mathop {mathop {15{O_2}}limits_{15(22.4l)}^{15mole} }limits_{'X'l} to 12C{O_2}(g) + 6{H_2}O(l)$

$large X = frac{39}{2 times 78} times 15(22.4)$

= 84 lit

Volume of O₂ required for the combustion of 39grams of C₆H₆ = 84 lit

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