Question

1. Suppose 0.7542 g of magnesium reacts with excess oxygen to form magnesium oxide as the only product, what would be the theoretical yield of the product?
2. If 0.8922 g of magnesium oxide is obtained from the reaction indicated in #1 above, what would be the percent yield of the magnesium oxide?




3. Suppose the percent yield was calculated to be over 100%, what possible reasons could you give to account for it?

Answer 1

Answer :

(1) The theoretical yield of product, $MgO$ is, 1.257 grams.

(2) The percent yield of $MgO$ is, 64.13 %

(3) If the percent yield is calculated to be over 100% then there might be some impurity present in the desired product.

Solution : Given,

Mass of Mg = 0.7542 g

Molar mass of Mg = 24 g/mole

Molar mass of MgO = 40 g/mole

First we have to calculate the moles of Mg.

$\text{ Moles of }Mg=\frac{\text{ Mass of }Mg}{\text{ Molar mass of }Mg}=\frac{0.7542g}{24g/mole}=0.03142moles$

Now we have to calculate the moles of $MgO$

The balanced chemical reaction is,

$2Mg(s)+O_2(g)\rightarrow 2MgO(s)$

From the reaction, we conclude that

As, 2 mole of $Mg$ react to give 2 mole of $MgO$

So, 0.03142 moles of $Mg$ react to give 0.03142 moles of $MgO$

Now we have to calculate the mass of $MgO$

$\text{ Mass of }MgO=\text{ Moles of }MgO\times \text{ Molar mass of }MgO$

$\text{ Mass of }MgO=(0.03142moles)\times (40g/mole)=1.257g$

Theoretical yield of $MgO$ = 1.257 g

Experimental yield of $MgO$ = 0.8922 g

Now we have to calculate the percent yield of $MgO$

$\% \text{ yield of }MgO=\frac{\text{ Experimental yield of }MgO}{\text{ Theretical yield of }MgO}\times 100$

$\% \text{ yield of }MgO=\frac{0.8922g}{1.257g}\times 100=70.97\%$

Therefore, the percent yield of $MgO$ is, 70.97 %

If the percent yield is calculated to be over 100% then the product would be greater than 1.257 g which indicates that there might be some impurity present in the desired product.