[carlsone]

0.144 g of an aluminium compound X react with an excess of water, to produce a gas. This gas burns completely in O₂ to form H₂O and 72 cm³ of CO₂ only. The volume of CO₂ was measured at room temperature and pressure.
What could be the formula of X?

[C = 12.0, Al = 27.0; 1 mole of any gas occupies 24 dm³ at room temperature and pressure]

A Al2C3  B Al 3C4  C Al 4C3  D Al 5C3

The 72cm3 is equivalent to 0.003 moles of Carbon. This is equal to 0.003 x 12 = 0.036g of Carbon. Subtracting from 0.144 g of the compound gives 0.108g of Aluminium or 0.108/27=0.004 moles. The ratio of the two moles gives C as the ans. I am not comfortable with this though,any more logically approaches.

There's another approach posted here:http://sgforums.com/forums/2297/topics/443584 but I am not following how the ratio of 4/3 was obtained, the expression is difficult to follow. Any help would be much appreciate for this first post .

[Borek]

The 72cm3 is equivalent to 0.003 moles of Carbon. This is equal to 0.003 x 12 = 0.036g of Carbon. Subtracting from 0.144 g of the compound gives 0.108g of Aluminium or 0.108/27=0.004 moles. The ratio of the two moles gives C as the ans. I am not comfortable with this though,any more logically approaches.

That's a perfectly correct approach, IMHO thebest one - you perfectly ignore everything that is irrelevant.

There's another approach posted here:http://sgforums.com/forums/2297/topics/443584 but I am not following how the ratio of 4/3 was obtained, the expression is difficult to follow. Any help would be much appreciate for this first post .

As you have correctly noticed, the approach presented there (while correct) is much more difficult to follow, which makes your answer better.

Basically they derived the equation

[tex]3\times 10^{-3} = y \frac {0.144}{27x + 12y}[/tex]

If you multiply both sides by 27x+12y, then group all terms containing x an one side of the equation and those containing y on the other side, you will get something like

[tex]ax=by[/tex]

which is equivalent to

[tex]\frac x y = \frac b a[/tex]

[carlsone]

This is appreciated thanks, it's good to have the alternative approach sorted out as well.