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 H2
O and 72 cm3
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