In the experiment it is found that 2.0769 gram of pure X produce 3. 67

1.	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
Answer» The balanced equation for the above reaction is 2 X + 5/2 O₂→ 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.

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

Answer»

The balanced equation for the above reaction is

2 X + 5/2 O₂→ 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.

Discussion

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