Q100 P - Chemistry: Molecular Nature Of Matter And Change

Question

In a combination reaction, 2.22 g of magnesium is heated with 3.75 g of nitrogen. 

(a) Which reactant is present in excess? 

(b) How many moles of product are formed? 

(c) After reaction, how many grams of each reactant and product are present?

Step-by-Step Solution

Answer

1. You need to write about the excess reactant
2. You need to calculate how many moles of product are formed.
3. You need to calculate grams of each reactant and product after reaction.

Step 1: Balanced equation

A combination reaction of magnesium heated with nitrogen forms lithium oxide. The balanced chemical equation is

$3Mg\left(s\right)+N_2\left(g\right)\to M{g}_3{N}_2\left(s\right)$

Step 2: Calculation of moles of Mg

According to the question,

Mass of Mg = 2.22g

 

Molecular mass of Mg = 24.305g/mol

 

Again you know,

$moles=\frac{mass\left(given\right)}{mass\left(molar\right)}$

Moles of Mg

$$\begin{gathered} =\frac{2.22}{24.305}\left(mol\right) \\ =0.0913\left(mol\right) \end{gathered}$$

Hence, moles of Mg are 0.0913mol.

Step 3: Calculation of moles of Mg 3 N 2 when Mg is limiting reagent

According to the balanced equation

$3Mg\left(s\right)+N_2\left(g\right)\to M{g}_3{N}_2\left(s\right)$

3 moles of Mg produces 1 mole Mg₃N₂

Now, moles of Mg₃N₂

$$\begin{gathered} =0.0913\times \frac{1}{3}\left(mol\right) \\ =0.0304\left(mol\right) \end{gathered}$$

Hence, moles of Mg₃N₂ when Mg is limiting reagent are 0.0304mol.

Step 4: Calculation of moles of N 2

According to the question,

Mass of N₂ = 3.75g

 

Molecular mass of N₂ = 28g/mol

 

Again you know,

$moles=\frac{mass\left(given\right)}{mass\left(molar\right)}$

Moles of N₂

$$\begin{gathered} =\frac{3.75}{28}\left(mol\right) \\ =0.1339\left(mol\right) \end{gathered}$$

Hence, moles of N₂ are 0.1339mol.

Step 5: Calculation of moles of Mg 3 N 2 when N 2 is limiting reagent

According to the balanced equation

$3Mg\left(s\right)+N_2\left(g\right)\to M{g}_3{N}_2\left(s\right)$

1 mole of N₂ produces 1 mole Mg₃N₂

Now, moles of Mg₃N₂

$$\begin{gathered} =0.1339\times \frac{1}{1}\left(mol\right) \\ =01339\left(mol\right) \end{gathered}$$

Hence, moles of Mg₃N₂ when N₂ is limiting reagent are 0.1339mol.

Step 6: Conclusion

Mg is the limiting reagent in this reaction as moles of Mg₃N₂ is less in terms of Mg.

Hence, N₂ is present in excess.

Step 7: Limiting reagent

In the given reaction

$3Mg\left(s\right)+N_2\left(g\right)\to M{g}_3{N}_2\left(s\right)$

Mg is the limiting reagent.

Step 8: Calculation of moles of Mg

According to the question,

 

Mass of Mg = 2.22g

 

Molecular mass of Mg = 24.305g/mol

 

Again you know,

$moles=\frac{mass\left(given\right)}{mass\left(molar\right)}$

Moles of Mg

$$\begin{gathered} =\frac{2.22}{24.305}\left(mol\right) \\ =0.0913\left(mol\right) \end{gathered}$$

Hence, moles of Mg are 0.0913mol.

Step 9: Calculation of moles of Mg 3 N 2 (product)

According to the balanced equation

$3Mg\left(s\right)+N_2\left(g\right)\to M{g}_3{N}_2\left(s\right)$

3 moles of Mg produces 1 mole Mg₃N₂

Now, moles of Mg₃N₂

$$\begin{gathered} =0.0913\times \frac{1}{3}\left(mol\right) \\ =0.0304\left(mol\right) \end{gathered}$$

Hence, moles of Mg₃N₂ when Mg is limiting reagent are 0.0304mol.

Step 10: Calculation of mass of Mg (reactant):

As Mg is the limiting reagent; so after reaction there will be no Mg.

Step 11: Calculation of moles of N 2 reacted

From the balanced equation, you can see

$3Mg\left(s\right)+N_2\left(g\right)\to M{g}_3{N}_2\left(s\right)$

3 moles of Mg reacts with 1 mole N_2.

Hence, moles of N₂ reacted

$$\begin{gathered} =0.0913\times \frac{1}{3}\left(mol\right) \\ =0.0304\left(mol\right) \end{gathered}$$

Hence, moles of N₂ reacted is 0.0304mol.

Step 12: Calculation of mass of N 2 reacted

Again you know,

mass = moles x mass (molar)

Molar mass of N₂ is 28g/mol.

Now, mass of N₂

$$\begin{gathered} =0.0304\times 28\left(g\right) \\ =0.8512\left(g\right) \end{gathered}$$

Hence, mass of N₂ reacted is 0.8512g.

Step 13: Calculation of mass of N 2 left (reactant)

Mass of N₂ left

$$\begin{gathered} =mass\left(initial\right)-mass\left(reacted\right) \\ =\left(3.75-0.8512\right)\left(g\right) \\ =2.8988\left(g\right) \end{gathered}$$

Hence, mass of N₂ left is 2.8988g.

Step 14: Calculation of mass of Mg 3 N 2 (product):

Again you know,

mass = moles x mass (molar)

Molar mass of Mg₃N₂ is 100.915g/mol.

Now, mass of Mg₃N₂

$$\begin{gathered} =0.0304\times 100.915\left(g\right) \\ =3.0678\left(g\right) \end{gathered}$$

Hence, mass of Mg₃N₂ is 3.0678g.

Step 15: Conclusion

After reaction, there are 0g Mg, 2.8988g N₂ and 3.0678g Mg₃N₂.