Question

10 g of magnesium is burnt with 0.56 g 0, in a
closed vessel. Which reactant is left in excess
and how much? (At wt. Mg = 24; O = 16)
(AIPMT-2014
(1) Mg. 0.16 g (2) 0₂ 0.16 g
(3) Mg, 0.44 g (4) 0%, 0.28 g
ismivad with 14 tres​

Answer 1

Answer:

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