Aluminum nitrite and ammonium chloride react to form aluminum chloride, nitrogen, and water. How many grams of each substance are present after 72.5 g of aluminum nitrite and 58.6 g of ammonium chloride react completely?

Explanation:

Given Data:

The mass of aluminium nitrite is 72.5 g

The mass of ammonium chloride is 58.6 g

The balanced chemical equation for the reaction is given as follows.

Al(NO2)3 + 3NH4Cl → AlCl3 + 3N2 + 6H2O

The number of moles can be determined by the formula given as follows.

Number of moles = Mass / Molar mass

The molar mass of aluminum nitrate and ammonium chloride is 164.998 g/mol and 53.49 g/mol respectively.

inserting the respective values in the formula given above.

Moles of Al(NO2)3 = 72.5 g / 164.998 g/mol = 0.439 mol

Moles of NH4Cl = 58.6 g / 53.49 g/mol = 1.096 mol

From the balanced equation,

3 moles of ammonium chloride requires 1 mole of aluminum nitrate.

So, 1 mole of ammonium chloride requires 1 / 3 mole of aluminum nitrate.

Thus, 1.096 mole of ammonium chloride will require (1 / 3) × 1.096 = 0.3653 mole of aluminum nitrite.

Here, the amount of aluminum nitrate is more than the required amount so ammonium chloride is the limiting reagent.

From the balanced chemical equation, 3 mole of ammonium chloride gives 1 mole of aluminum chloride.

So, 1.096 mole of ammonium chloride will give;

3 = 1

1.096  = x

x = (1.096 * 1 ) / 3 = 0.3653 mole of aluminum chloride.

Therefore, the number of moles of aluminum chloride is 0.3653 mol.

Since the molar mass of aluminum chloride is 133.34 g/mol

Substitute the respective values in the formula given above.

0.3653 mol = Mass / 133.34 g/mol

Mass = 0.3653 mol × 133.34 g/mol = 48.71 g

Therefore , the mass of aluminum chloride produces is 48.71 g.

The Ammonium chloride is completely used up in the reaction.

The amount of alumium nitrite used is =  Number of moles * Molar mass =  0.3653 * 164.998 = 60.27

Mass of alminium nitrite left = 72.5 - 60.27 = 12.23g