A volume of 50cm3 of sulphur (iv) oxide was passed over heated catalyst with 25cm3 of oxygen. At the end of the reaction, it was found that 50cm3 of a new oxide of sulphur had been formed , and none of the original gases remained. Work out the formula of the new oxide?
SO2 + O2 --> SxOy
50cc.....25cc.......50cc
Since all of these are gases we can equate volume with moles. 50 cc SO2 and 25 cc O2 means we have used twice as many moles SO2 as O2; therefore, the equation should be 2SO2 + O2 ==> 2SOy. What's y? It must add to 6 (4 from 2SO2 + 2 from O2 = 6) and y must be 3.
Proof: Convert to grams. 50 cc SO2 = (50/22,400)*64 = 0.1428 g SO2
25 cc O2 = (25/22,400)*32 = 0.0357
Total reactants = 0.1428 g + 0.0357 = 0.1785 grams.
Total products = (50/22,400)*80 = 0.1786 grams which agrees with the Law of mas Conservation.