Question

Magnesium will burn in air to form both Mg3N2 and MgO. What mass of each product would be found if burning a 3.26 g sample of magnesium to completion produces a combined total of 5.09 g of the two products

Answer 1

Answer: The mass of MgO formed is 3.502 grams and mass of $Mg_3N_2$ formed is 1.587 grams.

Explanation:

We are given:

Total mass of magnesium reacted = 3.26 g

Total mass of products formed = 5.09 g

To calculate number of moles of a substance, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$      .....(1)

• For calculating mass of MgO:

The chemical equation for the formation of magnesium oxide follows:

$2Mg+O_2\rightarrow 2MgO$

Let 'x' grams of magnesium is reacting to form magnesium oxide.

Molar mass of Magnesium = 24.3 g/mol

Putting values in equation 1, we get:

Number of moles of magnesium = $\frac{x}{24.3}mol$

By Stoichiometry of the reaction:

2 moles of magnesium produces 2 moles of MgO

So, $\frac{x}{24.3}$ moles of magnesium will produce $\frac{2}{2}\times \frac{x}{24.3}=\frac{x}{24.3}$ moles of MgO

Molar mass of MgO = 40.3 g/mol

Putting values in equation 1, we get:

Mass of MgO = $\frac{x}{24.3}\times 40.3=(1.66\times x)g$

Mass of magnesium oxide = $(1.66\times x)g$ = X grams

• For calculating mass of $Mg_3N_2$ :

The chemical equation for the formation of magnesium nitride follows:

$3Mg+N_2\rightarrow Mg_3N_2$

Mass of magnesium reacting = (3.26 - x)

Molar mass of Magnesium = 24.3 g/mol

Putting values in equation 1, we get:

Number of moles of magnesium = $\frac{(3.26-x)}{24.3}mol$

By Stoichiometry of the reaction:

3 moles of magnesium produces 1 moles of $Mg_3N_2$

So, $\frac{(3.26-x)}{24.3}$ moles of magnesium will produce $\frac{1}{3}\times \frac{(3.26-x)}{24.3}=\frac{(3.26-x)}{72.9}$ moles of $Mg_3N_2$

Molar mass of $Mg_3N_2$ = 101 g/mol

Putting values in equation 1, we get:

Mass of $Mg_3N_2=\frac{(3.26-x)}{72.9}\times 101=[(3.26-x)\times 1.38]g$

Mass of magnesium nitride = $[(3.26-x)\times 1.38]g$ = Y grams

• Calculating the mass of products:

Total mass of the products = 5.09 grams

$X+Y=5.09$

Putting values of 'X' and 'Y' in above equation, we get:

$(1.66\times x)+[(3.26-x)\times 1.38]=5.09\\\\x=2.11g$

Mass of MgO = $(1.66\times 2.11)=3.502g$

Mass of $Mg_3N_2=[(3.26-2.11)\times 1.38]=1.587g$

Hence, the mass of MgO formed is 3.502 grams and mass of $Mg_3N_2$ formed is 1.587 grams.