Question

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

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

Hence, option A is correct.