Combustion of which gas gives 7 moles of gaseous product from 6 moles of gaseous reactant?

Question

$1 .$ $1.$ $C H_{4};$ $CH_4;$
$2 .$ $2.$ $H_{2} C = C H_{2};$ $H_2C=CH_2;$
$3 .$ $3.$ $H C \equiv C H;$ $HC-=CH;$
$4 .$ $4.$ $H_{3} C - C H_{2} C H_{2}$ $H_3C-CH_2CH_2$

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Answer 1 · 2016-06-26T17:35:27

Propane combustion; $option 4$ $"option 4"$

$underbrace(C_3H_8(g) + 5O_2(g))_"6 moles" rarr underbrace(3CO_2(g) +4 H_2O(g))_"7 moles"$

Because $Vpropn" (T and P constant)"$ , the VOLUMES represent the stoichiometric equivalents of hydrocarbon and oxidant.

You should write out all the combustion reactions with each fuel to satisfy yourself that this is the case:

i.e. $C_2H_4(g) +3O_2(g) rarr 2CO_2(g) +2H_2O(g)$ , which stoichiometry disagrees with the experimental result.

You can do the rest.