102.1 g of Aluminum nitrite and 174.3 g of ammonium chloride react to form aluminum chloride, nitrogen, and water.

How many grams of the excess reagent remains after the reaction?

57.39 g excess Aluminum nitrite

Explanation:

When performing stoichiometric calculations, the first thing we need is the balanced chemical reaction.

In this case we will have:

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

(nitrite ion is NO₂⁻)

Now that we have the balanced reaction, we need to calculate the number of moles, n, of Al (NO₂) ₃ and NH₄Cl, and perform the calculations necessary to determine the excess reagent and its amount.

The number of moles is:

n = mass / MW where MW is the molecular weight and m the mass.

MW Al (NO₂) ₃ = 40.99 g/mol

MW NH₄Cl = 53.49 g/mol

n Al (NO₂) ₃ = 102.1 g / 40.99 g/mol = 2.49 mol Al (NO₂) ₃

n NH₄Cl = 174.3 g / 53.49 g/mol = 3.26 mol NH₄Cl

Now lets calculate how many moles of NH₄Cl will react with 2.49 mol Al (NO₂) ₃:

(3 mol NH₄Cl / 1 mol Al (NO₂) ₃) x 2.49 mol Al (NO₂) ₃ = 7.5 mol NH₄Cl

We only have 3.26 mol NH₄Cl. Therefore our limiting reagent is NH₄Cl, and the excess reagent is Al (NO₂) ₃

Now lets calculate the number of moles Al (NO₂) ₃ used to react with 3.26 mol NH₄Cl:

(1 mol Al (NO₂) ₃ / 3 mol NH₄Cl) x 3.26 mol NH₄Cl = 1.09 mol Al (NO₂) ₃

The excess number of moles is:

= 2.49 mol - 1.09 mol = 1.40 mol Al (NO₂) ₃

grams of Al (NO₂) ₃ in excess

1.40 mol Al (NO₂) ₃ x 40.99 g/mol = 57.39 g