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

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