Calculate the percentage yield of this reaction of 2MgO(s) if 0,15 moles of magnesium react with excess oxygen gas to form 5,4 grams of Magnesium oxide

Md Ariful Islam

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Given Moles of Mgo is 0.15 moles .

Mg react with O2 and form MgO .

2 mole of Mg gives 2 mole of Mgo .

So 0.15 moles of Mg Gives 0.15 moles of MgO

So , Mass Of MgO = 0.15 * 40 = 6 gm.

So , Maximum Mass of MgO = 6 gm . (Theoretical Mass ).

Experimental Mass of MgO = 5.4 gm .

% Yield (percentage yield) = ( Experimental Mass / Theoretical Mass ) * 100 .

 

percentage yield = (5.4 / 6) * 100 = 90 % .

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