A metalic element X with RAM x form two oxide p and q . Oxide p contain 70% of X and q contain 72.4% of X. If the first oxide has the formula X2O3 what is the formula of the second oxide?
Formula for first oxide = X2O3
Mass of metal X = x
% of metal in X2O3 = "\\frac{2x}{2x+48}" × 100
But as given % = 70%
x = 56
In second oxide
Metal given 72.4% , so oxygen will be 27.6 %
So the ratio is :
Metal : oxide
"\\frac{72.4}{56}" = "\\frac{27.6}{16}" =1.29:1.72 = 3:4
So formula for second oxide = X3O4 4
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