Question

Given reaction:

2NH3 (g) + 3Cl2 (g) → N2 (g) + 6HCl (g)
Determine the quantity in moles of the excess reactant in each of the following cases. Calculate the maximum number of moles of HCl for each case for which the amount of reactants are given below.
a)1mol of NH3 and 1.5 mol of Cl2
b)6x 1022 NH3 molecules and 21x 1022 Cl2 molecules c)10.2 g of NH3 and 134.9g of Cl2
d)1.12 L of NH3 at STP and 4.48 L of Cl2 at STP

Answer 1

To determine the quantity in moles of the excess reactant, we need to find the limiting reactant first. The excess reactant is the reactant that is not limiting and has leftover moles after the reaction is complete. In each case, we compare the moles of each reactant to the stoichiometric ratio in the balanced equation to determine the limiting and excess reactants.

Step-by-step explanation:

To determine the quantity in moles of the excess reactant, we need to find the limiting reactant first. To do this, we compare the ratio of the moles of each reactant to the stoichiometric ratio in the balanced equation. Whichever reactant gives the smaller mole ratio is the limiting reactant. The excess reactant is the reactant that is not limiting and has leftover moles after the reaction is complete.

a) To find the excess reactant, we compare the moles of NH3 and Cl2. From the balanced equation, we know that 2 moles of NH3 react with 3 moles of Cl2. Using the given amounts, we have:

NH3:Cl2 = 1 mol:1.5 mol = 2 mol:3 mol. Since the mole ratio of NH3 to Cl2 is higher than the stoichiometric ratio of 2:3, NH3 is the limiting reactant and Cl2 is the excess reactant.

b) To find the excess reactant, we compare the number of molecules of NH3 and Cl2. From the balanced equation, we know that 2 moles of NH3 react with 3 moles of Cl2. Using Avogadro's number, we can convert the given amounts to moles:

NH3:Cl2 = 6x 10^22 molecules:21x 10^22 molecules. Since the mole ratio of NH3 to Cl2 is higher than the stoichiometric ratio of 2:3, NH3 is the limiting reactant and Cl2 is the excess reactant.

c) To find the excess reactant, we compare the moles of NH3 and Cl2. From the balanced equation, we know that 2 moles of NH3 react with 3 moles of Cl2. Using the molar mass of NH3 (17 g/mol) and Cl2 (71 g/mol), we can convert the given masses to moles:

NH3:Cl2 = 10.2 g:134.9 g = 0.6 mol:1.9 mol. Since the mole ratio of NH3 to Cl2 is higher than the stoichiometric ratio of 2:3, NH3 is the limiting reactant and Cl2 is the excess reactant.

d) To find the excess reactant, we compare the volumes of NH3 and Cl2 at STP. From the balanced equation, we know that 2 moles of NH3 react with 3 moles of Cl2. Using the molar volume of a gas at STP (22.4 L/mol), we can convert the given volumes to moles:

NH3:Cl2 = 1.12 L:4.48 L = 1 mol:4 mol. Since the mole ratio of NH3 to Cl2 is lower than the stoichiometric ratio, NH3 is the excess reactant and Cl2 is the limiting reactant.