Question

To determine the moles of ammonium chloride produced, we need to find the limiting reactant. H2O is present in an unlimited amount, so it cannot be the limiting reactant. Let’s then determine the amount of each remaining reactant present (in moles). NaCl is already given in moles (0.250 mol). NH3 and CO2 are not. They are given in liters. How many moles are present in 3.08 L of NH3 and 2.09 L of CO2?

Answer 1

0.094 mols of CO2 are present in 2.09 L.

0.132 mols of NH3 are present in 3.08 L

The moles of ammonium chloride produced are 0.094

Step-by-step explanation:

First of all, think the reaction

NH3 + 2H2O + NaCl + CO2 → NH4Cl + NaOH + H2CO3

To get the moles with volume, you must need density

Density = mass / volume

Density CO2 = 0,001976 g/ml

Density NH3 = 0,00073 g/ml

Volume of CO2 = 2.09L

2.09 L . 1000 = 2090 mL

0,001976 g/ml = mass / 2090 mL

4.13 g = mass

Molar weight = 44 g/m

mass / molar weight = 4.13 g / 44 g/m = 0.094 moles

Volume of NH3 = 3.08 L

3.08 L . 1000 = 3080 mL

0,00073 g/ml = mass/ 3080 mL

2.25 g = mass

Molar weight = 17 g/m

mass / molar weight = mols → 2.25 g / 17g/m = 0.132 mols

In my reagents, the least amount I have is CO2. This is my limiting reactant. Take account the reaction.

1 mol of NH3 reacts with 1 mol of CO2, so as I have 0.132 moles of NH3 I need 0.132 moles of CO2; I only have 0.094.

Relation between CO2 and NH4Cl is 1:1 so, 1 mol of CO2 is needed to produce 1 mol e chloride, so 0.094 mols are needed to produce the same amount of chloride.