Question

During a synthesis reaction, 3.2 grams of magnesium reacted with 12.0 grams of oxygen. What is the maximum amount of magnesium oxide that can be produced during the reaction? Mg + O2 → MgO 5.3 grams 6.5 grams 7.2 grams 9.5 grams

Answer 1

Answer : The maximum amount of magnesium oxide produced can be, 5.3 grams.

Solution : Given,

Mass of Mg = 3.2 g

Mass of $O_2$ = 12.0 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 $O_2$.

$\text{ Moles of }Mg=\frac{\text{ Mass of }Mg}{\text{ Molar mass of }Mg}=\frac{3.2g}{24g/mole}=0.133moles$

$\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{12.0g}{32g/mole}=0.375moles$

Now we have to calculate the limiting and excess reagent.

Synthesis reaction : It is defined as the reaction in which the two smaller molecule combined to give one larger molecule.

The balanced chemical reaction is,

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

From the balanced reaction we conclude that

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

So, 0.133 moles of $Mg$ react with $\frac{0.133}{2}=0.0665$ 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, 0.133 moles of $Mg$ react to give 0.133 moles of $MgO$

Now we have to calculate the mass of $MgO$

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

$\text{ Mass of }MgO=(0.133moles)\times (40g/mole)=5.3g$

Therefore, the maximum amount of magnesium oxide produced can be 5.3 grams.

Answer 2

Answer : The correct option is, 5.3 grams

Explanation :

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

$\text{Moles of }Mg=\frac{\text{Mass of }Mg}{\text{Molar mass of }Mg}=\frac{3.2g}{24g/mole}=0.133moles$

$\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{12g}{32g/mole}=0.38moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$2Mg+O_2\rightarrow 2MgO$

From the balanced reaction we conclude that

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

So, 0.133 moles of $Mg$ react with $\frac{0.133}{2}=0.066$ moles of $O_2$

That means, in the given balanced reaction, $Mg$ is a limiting reagent because it limits the formation of products and $O_2$ is an excess reagent.

Now we have to calculate the moles of MgO.

As, 2 moles of $Mg$ react with 2 moles of $MgO$

So, 0.133 moles of $Mg$ react with $\frac{2}{2}\times 0.133=0.133$ moles of $MgO$

Now we have to calculate the mass of MgO.

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

$\text{Mass of }MgO=(0.133mole)\times (40.3g/mole)=5.3g$

Therefore, the mass maximum amount of magnesium oxide is, 5.3 grams