Stoichiometry of an Acid–Base Reaction Experiment 1: Prepare a Sodium Carbonate

Question

Stoichiometry of an Acid–Base Reaction

Experiment 1: Prepare a Sodium Carbonate Solution

Lab Results

Record the following masses.

Data Analysis

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

Experiment 2: Neutralization Reaction

Lab Results

What was the acid in the reaction?

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:

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?

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

Explanation / Answer

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.

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