HKCEE Chemistry ( help)

1. Metal X forms an oxide. 27.53g of this oxide contain 24.96g of X. What is the mole ratio of X to oxygen in the oxide? (relative atomic masses: O = 16.0 , X = 207.0) (HKCEE 2000)

A 1:1
B 1:2
C 2:3
D 3:4
ANSWER
mole of x
=24.96/207
mole of oxygen
=(27.53-24.96)/16
=2.57/16
the mole ratio of x to oxygen in the oxide
=24.96/207 :2.57/16
=399.36: 531.99
=0.7507
=3/4
So the answer is D
2.gases x and y react to give a gaseous product z.the reaction can be represented by the equation:x(g) + 3y(g)------>2 z(g).in an experiment,40 cm*3 of x and 60 cm*3 of y are mixed and allowed to react in a closed vessel .what is the volume of the resultant gaseous mixture (all volumes are measured at room temperature and pressure )
ANSWER
From the equation given in the question, we can see that:
1 moles of gas X react with 3 moles of gas Y to yield 2 moles of gas Z.
That is 1 cm^3 of gas X react with 3 cm^3 of gas Y to yield 2 cm^3 of gas Z.
Similarly 20 cm^3 of gas X react with 60 cm^3 of gas Y to yield 40 cm^3 of gas Z.
So in the final stage:
Volume of gas X = 40-20=20 cm3
Volume of gas Y = 0 cm3
Volume of gas Z = 40 cm3
Therefore total volume of gas mixture in the final stage = 20 + 40 = 60 cm3.

(1) In 27.53 g of the metal oxide, there are:

24.96 g of metal X → 24.96/207 = 0.121 moles
27.53 - 24.96 = 2.57 g of O → 2.57/16 = 0.161 moles
So we can see that the mole ratio of X to O is approximately 3:4.
(2) First of all, by Avogadro's law, we can say that the initial mole ratio of gases X and Y is:
X : Y = 2 : 3
And from the equation, 1 mole of X reacts with 3 moles of Y.
Therefore X is in excess in which:

Volume of gas X remained = 20 cm3
Volume of gas Y remained = 0 cm3
Volume of gas Z produced = 40 cm3
Hence volume of the resultant gaseous mixture = 20 + 40 = 60 cm3 .