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21 Chemistry -- Stoichiometry

2g of magnesium is burnt in a closed vessel containing 3g of oxygen.                                      i. Which one is limiting reactant?                  ii. Calculate the mole of reactant left over. iii. How many gram of MgO are produced? iv. What is the mass of H2SO4 required to n...

2g of magnesium is burnt in a closed vessel containing 3g of oxygen.                                     
 i. Which one is limiting reactant?                 
ii. Calculate the mole of reactant left over.
iii. How many gram of MgO are produced?
 iv. What is the mass of H2SO4 required to neutralize MgO formed in the reaction?

1 Answers

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Solution:                                                        
The required balance chemical equation:

              2Mg + O2 ---------------> 2MgO

i) For limiting reactant.                                   
   According to the balanced chemical equation.

   48g Mg reacts with 32g of O2 
   1g Mg reacts with (32 ÷ 48)g of O2 
   2g Mg reacts with 0.67 × 2g O2 

  But the mass of O2 given is 3g. This means that Mg is finished first in the reaction. So, Mg is limiting reactant. 

ii.
   Given mass of O2 = 3g
   Consumed mass of O2 = 1.33g 

   Therefore,
   No. of O2 left over = (Given mass -- Consumed mass)/molecular mass of O2 
                                    = (3 - 1.33)/32 = 0.052 mole

iii.
   According to the balanced chemical equation:
    48g Mg produces 80g MgO 
    1g Mg produces ( 80 ÷ 48)g MgO
    2g Mg produces 1.67 × 2g MgO
                                  = 3.33g MgO

iv.
   The balanced chemical equation of MgO with H2SO4 is given below 

            MgO + H2SO4 ----------------> MgSO4 + H2O 

    Acc. to the balanced chemical equation:
    40g MgO reacts with 98g H2SO4 
    1g MgO reacts with (98/40)g H2SO4
                                      = 2.45g H2SO4

    We have, MgO = 3.33g
                     3.33g  MgO reacts with 2.45 x 3.33g H2SO4
                                                               = 8.16g H2SO4

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