Question

Stoichiometry of an Acid–Base Reaction

Experiment 1: Prepare a Sodium Carbonate Solution

Lab Results

Record the following masses.

a	mass of empty beaker (g)	85.000 g
b	mass of beaker plus Na2CO3 (g)	87.000 g
c	mass of Na2CO3 (g)	2.206 g

Data Analysis

Convert the mass of Na₂CO₃ to moles, given its molar mass (MM) of 105.989 g/mol.

2.206 g / 105.989 g/mol = 0.0208 mol

Experiment 2: Neutralization Reaction

Lab Results

What was the acid in the reaction?

HCl

Data Analysis

1. How many moles of HCl did you add to fully neutralize the Na₂CO₃ solution?

Experiment 3: Isolate Sodium Chloride

Lab Results

Record the following masses:

a	mass of empty beaker (g)	85.000 g
b	madd of beaker plus NaCl (after boiling off the water) (g)	87.206 g

Data Analysis

2. Convert the mass of NaCl to moles, given its molar mass (MM) of 58.443 g/mol.

Conclusions

3. Calculate the experimentally determined molar ratio of Na₂CO₃ to NaCl using the formula below.
molar ratio = (mol Na₂CO₃) / (mol NaCl)

4. Use the theoretical molar ratio to calculate the theoretical yield of NaCl in grams from 2.000 g of Na₂CO₃?

5. The percent yield is the ratio of the actual amount of a product to the theoretical amount. Calculate the percent yield of NaCl as shown below.
%yield = (experimental yield) / (theoretical yield) × 100

6. Given the data below, how many grams of CO₂ would you expect to be formed in the reaction of excess HCl with the Na₂CO₃?

mass of empty beaker (g)	84.000
mass of beaker plus Na2CO3 (g)	86.375
mass of Na2CO3 (g)	2.375

Answer 1

The neutralization Reaction between HCl and Na₂CO₃ is:

2HCl + Na₂CO₃ ---------→ 2NaCl + H₂O + CO₂

Question-1

2 moles of HCl needed 1 mole of Na₂CO₃

therefore, 0.0208 moles of Na₂CO₃ require, 2x 0.0208 = 0.0416 moles of HCl

Question-2

mass of NaCl = Mass of beaker plus NaCl – mass of beaker = 87.206 g- 85.000 g =2.206 g of NaCl

Now, Number of moles of NaCl = Given mass/ molar mass

= 2.0206 / 58.443 =0.03457 moles of NaCl

Question-3

Molar ratio = Moles of Na₂CO₃/ Moles of NaCl = 0.0208 / 0.03457 =0.6016

Question-4

The theoretical molar ratio of NaCl from 2 g of Na₂CO₃;

Generally, 1 mole of Na₂CO₃ will produce 2 moles of NaCl

So the molar ratio is = ½ or 0.5

Question-5

in terms of yield, 106 g of Na₂CO₃ will give 2x 58.443 = 116.886 g of NaCl

So 2 g of Na₂CO₃ will give , (116.886 /106)x 2 =2.2053 g of NaCl

Therefore yield, = (Experimental yield / theoretical yield ) x 100 = 2.206/2.2053 = 100.03% yield.

Question-6

Generally from the neutralization reaction, 106 g of Na₂CO₃ will give 44 g of CO₂

So 2.375 g of Na₂CO₃ will give = (44/106) x 2.375 =0.9858g of CO₂ will obtain.