Answer:
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
