Ammonia gas combines with hydrogen chloride gas, forming solid ammonium chloride. a. Write a balanced chemical equation for this reaction. b. In a reaction mixture of 3.0 g ammonia and 5.0 g hydrogen chloride, which of the two is the limiting reagent? c. How many grams of ammonium chloride could form from the reaction mixture in part b? d. How much of the reactant is left over in the reaction mixture o

a. NH₃ (g) + HCl (g) → NH₄Cl (s)

b. HCl is the limiting reagent

c. 7.35 g of NH₄Cl

d. 0.039 moles of NH₃ remains after the reaction is complete (0.663 grams of ammonia)

Explanation:

First of all, we define the reaction where the reactants are the ammonia and the hydrogen chloride. The product is the ammonium chloride.

NH₃ (g) + HCl (g) → NH₄Cl (s)

First of all, we need to determine the limiting reactant. As the ratio is 1:1, the compound that has the lowest mol, will be the limiting.

3 g / 17g/mol = 0.176 moles NH₃

5 g / 36.45 g/mol = 0.137 moles HCl (limiting reactant)

For 0.176 moles of ammonia, we need the same of HCl, but we have 0.137 moles. We convert the moles to mass, in order to know how many grams of NH₄Cl are produced → 0.137 mol. 53.45 g / 1mol = 7.32 g

As ratio is 1:1, and the limiting reagent is the HCl, after the reaction goes complete, some ammonia remains. This is because, the ammonia is the excess reagent.

0.176 mol - 0.137 moles = 0.039 moles of NH₃ remains after the reaction is complete; we convert the moles to mass → 0.039 mol. 17 g / mol = 0.663 g are left over in the reaction mixture.