Answer to Question #124150 in Organic Chemistry for Lethabo Mohlaloga

Question

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

Answer 1

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 % .