If the percentage yield of this reaction is 80%, calculate the mass of magnesium that needs to be burned to produce 30g of magnesium oxide

Answer 1

Percentage yield is defined as actual yield / theoretical yield (x100%).

We know the actual yield is 30 g of magnesium oxide (MgO), and we know the yield is 80% (0.80). So using this information, AND THE CORRECTLY BALANCED EQUATION, we can find the mass of Mg needed.

Burning Mg:

2Mg + O₂ ==> 2MgO ... balanced equation

molar mass MgO = 40.30 g/mol

30 g MgO x 1 mol / 40.30 g = 0.744 mols MgO

From the balanced equation we can convert this to mols of Mg needed:

0.744 mols MgO x 2 mols Mg / 2 mols MgO = 0.744 mols Mg needed, if the yield were 100%

Since yield is only 80%, we need 20% more to account for this deficiency:

0.744 mols x 1.20 = 0.893 mols Mg needed @ 80% yield to obtain 0.744 mols MgO

Finally, we convert this amount of Mg to grams:

0.893 mols Mg x 24.3 g Mg/mol = 21.7 g Mg = 20 g Mg (1 sig. fig. based on 1 sig. fig. in 30 g given in question)