Liquid benzene burns in oxygen according to2C6H6 + 15O2=12CO2 +6H2O . how many litres of O2 at STP are needed to complete the combustion of 39g of liquid benzene
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Answered by
67
by looking at the chemical equation it is certain that,
2 molecules of benzene react with 15 molecules of oxygen to produce 12 & 6 molecules of carbon dioxide and water respectively.
hence,
78*2 g of benzene is required to burn= 15*22.4 litre of oxygen at STP.
156 grams of benzene is required to burn 336 litres of oxygen at STP.
so, according to the question,
39g of benzene will be required to burn =(336/156)*39 litres of oxygen at STP.
=84 litres of oxygen at STP.
That is the answer.
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2 molecules of benzene react with 15 molecules of oxygen to produce 12 & 6 molecules of carbon dioxide and water respectively.
hence,
78*2 g of benzene is required to burn= 15*22.4 litre of oxygen at STP.
156 grams of benzene is required to burn 336 litres of oxygen at STP.
so, according to the question,
39g of benzene will be required to burn =(336/156)*39 litres of oxygen at STP.
=84 litres of oxygen at STP.
That is the answer.
If you still have doubts or if you're stuck at a sum then you could try asking Expert Online Chemistry Tutors @https://tutstu.com/tutors/Chemistry/sub/15/
Take a free trial with the Tutor you choose and get solutions to any problems that you might be facing. All you need to do is to Register as Student for Free @https://user.tutstu.com/student-login-register.
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