Question #9140f

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Question #9140f

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Answer 1 · 2015-05-28T01:55:25

37.5 litres of #C_2H_2# is required and 37.5 litres of water vapour is produced.

#2C_2H_(2(g))+5O_(2(g))rarr4CO_(2(g))+2H_2O_((g))#

From the equation you can see that if 75.0L of #CO_2# is produced then 75/2 = 37.5L of #C_2H_2# is required since 1 mole #C_2H_2# produces 2 mole #CO_2#.

2 moles #C_2H_2# produces 2 moles #H_2O#.

So 37.5L #C_2H_2# produces 37.5L #H_2O# vapour.

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Question #9140f

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