Question

1g of magnesium is burnt with 0.56g of O2 in a closed vessel.Which reactant is left in excess and how much?​

Answer 1

Answer:

Which reactant is left in excess and how much? From the above equation, it is clear that, 24gMg reacts with 16gO2. But only 0.56gO2 is available which is less than 0.67g. Thus, O2 is the limiting reagent.

Explanation:

The reaction involved is:

2Mg+O

2

⟶2MgO

Now, No. of moles of Mg =

2.4g/mol

1.0g

=0.0416 moles

No. of moles of O

2

=

32g/mol

0.56g

=0.0175 moles

1 mole O

2

consumes 2 mole Mg & gives 2 mole MgO

⇒O

2

here is limiting reagent as 0.0416 moles of Mg require 0.0208 mole of O

2

but only 0.0175 moles of O

2

given.

⇒0.0175 moles O

2

will consume 2×0.0175 moles of Mg

∴ No. of moles of Mg consumed = 0.035 moles

⇒ Remaining moles of Mg=0.0416−0.035

=0.0066 moles of Mg

As we know, 1 mole of Mg=24g

⇒0.0066 moles of Mg=24×0.006=0.1584g

∴ Amount of Mg left in excess = 0.16g