Question

How many grams of magnesium oxide can be produced if 25.0 g of magnesium are allowed to react with 28 g of oxygen?

Answer 1

Answer : The mass of $MgO$ produced is, 41.68 grams.

Explanation : Given,

Mass of $Mg$ = 25.0 g

Mass of $O_2$ = 28 g

Molar mass of $Mg$ = 24 g/mol

Molar mass of $O_2$ = 32  g/mol

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

$\text{Moles of }Mg=\frac{\text{Given mass }Mg}{\text{Molar mass }Mg}$

$\text{Moles of }Mg=\frac{25.0g}{24g/mol}=1.042mol$

and,

$\text{Moles of }O_2=\frac{\text{Given mass }O_2}{\text{Molar mass }O_2}$

$\text{Moles of }O_2=\frac{28g}{32g/mol}=0.875mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

$2Mg+O_2\rightarrow 2MgO$

From the balanced reaction we conclude that

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

So, 1.042 moles of $Mg$ react with $\frac{1.042}{2}=0.521$ moles of $O_2$

From this we conclude that, $O_2$ is an excess reagent because the given moles are greater than the required moles and $Mg$ 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, 2 mole of $Mg$ react to give 2 mole of $MgO$

So, 1.042 mole of $Mg$ react to give 1.042 mole of $MgO$

Now we have to calculate the mass of $MgO$

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

Molar mass of $MgO$ = 40 g/mole

$\text{ Mass of }MgO=(1.042moles)\times (40g/mole)=41.68g$

Therefore, the mass of $MgO$ produced is, 41.68 grams.