Question

During a synthesis reaction, 2.4 grams of magnesium reacted with 8.0 grams of oxygen. What is the maximum amount of magnesium oxide that can be produced during the reaction?
Mg + O2 → MgO


2.1 grams

2.8 grams

3.6 grams

3.9 grams

Answer 1

Answer: The maximum amount of magnesium oxide produced during the reaction can be, 3.9 grams.

Explanation : Given,

Mass of $Mg$ = 2.4 g

Mass of $O_2$ = 8.0 g

Molar mass of $Mg$ = 24.3 g/mol

Molar mass of $O_2$ = 31.9 g/mol

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

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

$\text{Moles of }Mg=\frac{2.4g}{24.3g/mol}=0.099mol$

and,

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

$\text{Moles of }O_2=\frac{8.0g}{31.9g/mol}=0.25mol$

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, 0.099 moles of $Mg$ react with $\frac{0.099}{2}=0.049$ 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, 0.099 mole of $Mg$ react to give 0.099 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 $MgO$ = 40.3 g/mole

$\text{ Mass of }MgO=(0.099moles)\times (40.3g/mole)=3.9g$

Therefore, the maximum amount of magnesium oxide produced during the reaction can be, 3.9 grams.