Question

Magnesium oxide can be made by heating magnesium metal in the presence of the oxygen. The balanced equation for the reaction is 2 Mg(s) + O2(g) → 2 MgO(s) Now consider that you react 10.0 g Mg with 6.00 g O2 gas. If you were able to collect 9.62 g of MgO, what would be your percent yield for the reaction?

Answer 1

Answer : The percent yield of $MgO$ is, 64.13 %

Solution : Given,

Mass of Mg  = 10 g

Mass of $O_2$ = 6 g

Molar mass of Mg = 24 g/mole

Molar mass of $O_2$ = 32 g/mole

Molar mass of MgO = 40 g/mole

First we have to calculate the moles of Mg and $O_2$.

$\text{ Moles of }Mg=\frac{\text{ Mass of }Mg}{\text{ Molar mass of }Mg}=\frac{10g}{24g/mole}=0.4167moles$

$\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{6g}{32g/mole}=0.1875moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

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

From the balanced reaction we conclude that

As, 1 mole of $O_2$ react with 2 mole of $Mg$

So, 0.1875 moles of $O_2$ react with $0.1875\times 2=0.375$ moles of $Mg$

From this we conclude that, $Mg$ is an excess reagent because the given moles are greater than the required moles and $O_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $MgO$

From the reaction, we conclude that

As, 1 mole of $O_2$ react to give 2 mole of $MgO$

So, 0.1875 moles of $O_2$ react to give $0.1875\times 2=0.375$ moles of $MgO$

Now we have to calculate the mass of $MgO$

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

$\text{ Mass of }MgO=(0.375moles)\times (40g/mole)=15g$

Theoretical yield of $MgO$ = 15 g

Experimental yield of $MgO$ = 9.62 g

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

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

$\% \text{ yield of }MgO=\frac{9.62g}{15g}\times 100=64.13\%$

Therefore, the percent yield of $MgO$ is, 64.13 %