0.5400 g of a metal X yields 1.020 g of its oxide X2O3. The number of ...
We are given that, 0.54 g of X gives 1.020 g of its oxide. Therefore, according to the law of conservation of mass,
Amount of Oxygen used by 0.54 g of X to form its oxide = (1.020-0.54)g = 0.48 g of Oxygen.
We know that, 32 g of O2 = 1 mole of O2.
So, 0.48 g of O2 = (0.48/32) = 0.015 mole of O2.
From the equation, 0.48 g of O2 reacts with 0.54 g of X.
Therefore, (3*32) g of O2 will react with (3*32*0.54)/0.48 = 108 g of X.
If 108 g of X = 4 mole of X.
So, 0.54 g of X = (0.54*4)/108 = 0.02 mole of X.
Thus, the correct answer is (B) 0.02 mole of X are present.