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=(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=(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
`(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* \text{ Molar mass of }MgO`

Molar mass of
`MgO` = 40.3 g/mole

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

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