In the experiment it is found that 2.0769 gram of pure X produce 3. 6769 gram of pure x2O5 the number of mole of X is


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The balanced equation for the above reaction is 

2 X + 5/2 O→ X2O5

Now the mass of oxygen using law of conservation of mass = 3.6769 - 2.0769

Hence mass of Oxygen utilized in the reaction = 1.6 g 

number of moles of oxygen = given mass / molar mass 

Hence the number of moles of oxygen = 1.6/ 32 = 0.05 moles 

As 5/2 moles of O2 reacts with 2 moles of X

Hence 

5/2 mole of O2 → 2 moles of X

1 mole of O2 → 4/5 moles of X

0.05 moles of O2 reacts with → 4× 0.05/ 5 moles of X

Hence the number of moles of  X present in the system = 0.04.