STOICHIOMETRY: QUANTITATIVE INFORMATION ABOUT CHEMICAL REACTIONS

Question

Ammonia gas can be prepared by the reaction of a metal oxide such as calcium oxide with ammonium chloride.

CaO(s) + 2NH₄CI(s) --> 2NH₃(g) + H₂O(g) + CaCl₂(s)

If 269g of CaO and 415g of NH₄Cl  are mixed, what is the limiting reactant? CaO ? NH₄Cl ? NH₃ ?

_________

What mass of NH₃  can be produced?

_______g

Answer 1

The limiting reactant is the reactant that yields the least amount of product in a reaction. Therefore, you need to calculate how much product is yielded from each reactant and compare the two values.

Consider the reaction:

CaO(s) + 2NH₄CI(s) --> 2NH₃(g) + H₂O(g) + CaCl₂(s)

The molecules on the left of the reaction arrow, CaO and NH₄Cl, are the reactants. The molecules on the right of the reaction arrow, NH₃, H₂O, and CaCl₂, are the products. According to this reaction equation, it takes one mole of CaO and two moles of NH₄Cl to make two moles of NH₃.

You can find the molar mass of a molecule by adding up the atomic masses of each atom making up that molecule. You can find the atomic masses of each element from the periodic table.

Molar mass of CaO: (40.078) + (15.999) = 56.077 g/mol

Molar mass of NH₄Cl: (14.007) + 4(1.008) + 35.45 = 53.489 g/mol

It is also important to know the mole ratios to complete your calculations.

1 mol CaO/2 mol NH₃

2 mol NH₄Cl/2 mol NH₃ = 1 mol NH₄Cl/1 mol NH₃

Now we multiply such that the units we're converting from are cancelled out.

How many moles of NH3 do we get from 269 g CaO?

269 g CaO x (1 mol CaO/56.077 g CaO) x (2 mol NH3/1 mol CaO) = 9.5939 mol NH3.

How many moles of NH3 do we get from 415 g NH4Cl?

415 g NH4Cl x (1 mol NH4Cl/53.489 g NH4Cl) x 1 mol NH3/1 mol NH4Cl) = 7.7586 mol NH3.

You get fewer moles of NH3 from your given amount of NH4Cl. Therefore, NH4Cl is the limiting reactant.

All that's left to do is convert your moles of NH3 to grams.

We get 7.7586 mol NH3 from the reaction.

The molar mass of NH3 is (14.007) + 3(1.008) = 17.031 g/mol.

7.7586 mol NH3 x (17.031 g NH3/1 mol NH3) = 132.1367 g NH3.