Question

0.5400g of a metal X yields 1.020g of its oxideX2O3.The no. of moles of X is:- (A)0.01 (B)0.02 (C)0.04 (D)0.05

Answer 1

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, (3x32) g of O2 will react with (3x32x0.54)/0.48 = 108 g of X.

If 108 g of X = 4 mole of X.

So, 0.54 g of X = (0.54x4)/108 = 0.02 mole of X.

Thus, the correct answer is (B) 0.02 mole of X are present.