Question

Magnesium oxide can be made by heating magnesium metal in the presence of oxygen. The balanced equation for the reaction is:
2Mg(s)+O2(g)→2MgO(s) When 10.1 g of Mg are allowed to react with 10.5 g of O2, 13.9 g of MgO are collected.

a) Determine the limiting reactant for the reaction.

b) Determine the theoretical yield for the reaction.

c) Determine percent yield for the reaction.

Answer 1

Given reaction:

2 Mg(s)+O₂(g) → 2 MgO(s)

PART (a):

No.of moles of Mg = (10.1 g) / (24.3 g/mol) = 0.416 mol

No.of moles of O₂= (10.5 g) / (32 g/mol) = 0.328 mol

From the balanced reaction:

2 moles of Mg requires 1 mole of O₂.

1 mole of Mg requires 1/2 mole of O₂.

Here, we have 0.416 mole of Mg which will only require (0.416/2) mole of O₂ = 0.208 mole of O₂

Therefore, We have excess of O₂ and Mg will be consumed completely in the reaction.

So the limiting reagent is Mg.

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PART (b):

Since Mg is the limiting reagent, we need to do calculations based on amount Mg.

From the balanced reaction, 2 moles of Mg reacts to form 2 moles of MgO.

Hence, 1 mole of Mg will produce 1 mole of MgO.

Hence, We have 0.416 mole of Mg which will react to produce 0.416 mole of MgO.

Molar mass of MgO = 40.3​​​​​​​ g/mol

Hence, Theoretical yield of MgO = (0.416 mol) x (40.3 g/mol) = 16.75 g

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PART (c):

Actual yield of MgO = 13.9 g

Therefore, Percentage yield = (actual yield) / (theoretical yield) x 100%

= (13.9 g)/(16.75 g) x 100 %

= 0.83 x 100%

= 83 %

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