Question

9) In a closed vessel, 0.56 g of
oxygen is burnt with the 1.0 g
of magnesium. which reactant
is left in excess and how much
?
1) 02,0.28 g
2) Mg, 0.16g
3) Mg, 0.44 g
4) 02,0.16 g​

Answer 1

Answer:

2) Mg, 0.16g

Explanation:

The reaction involved is:

2Mg+O2 ⟶2MgO

Now, No. of moles of Mg =  1.0g/2.4g/mol =  =0.0416 moles

No. of moles of O2=  0.56g/32g/mol =  =0.0175 moles

1 mole O2 consumes 2 mole Mg & gives 2 mole MgO

​  

 ⇒ O2 here is limiting reagent as 0.0416 moles of Mg require 0.0208 mole of O2 but only 0.0175 moles of O2 is given.

⇒0.0175 moles O2 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