Question

Question 10

(05.06 HC)
A 3.81-gram sample of NaHCO3 was completely decomposed in an experiment.
2NaHCO3-Na₂CO3 + H₂CO3
In this experiment, carbon dioxide and water vapors combine to form H₂CO3. After decomposition, the Na₂CO3 had
a mass of 2.86 grams.
1. Determine the mass of the H₂CO3 produced.
2. Calculate the percentage yield of H₂CO3 for the reaction. Show your work or describe the calculation
process in detail,
(10 points)

Answer 1

1. the mass of H2CO3 produced is 1.675 grams.

2. the percentage yield of H2CO3 is 59.7%.

Step-by-step explanation:

1. Using stoichiometry, we can find the moles of NaHCO3 and Na2CO3 produced in the reaction:

2 NaHCO3 -> Na2CO3 + H2CO3

Moles of NaHCO3 = 3.81 g / 84.01 g/mol = 0.04532 mol

Moles of Na2CO3 = 2.86 g / 105.99 g/mol = 0.02700 mol

Since the reaction produces an equal amount of Na2CO3 and H2CO3, we know that 0.02700 mol of H2CO3 was produced.

To find the mass of H2CO3 produced, we can use its molar mass:

Mass of H2CO3 = 0.02700 mol x 62.03 g/mol = 1.675 g

Therefore, the mass of H2CO3 produced is 1.675 grams.

2. The theoretical yield of H2CO3 is the amount that would be produced if the reaction went to completion and all of the NaHCO3 was converted to Na2CO3 and H2CO3. We can calculate the theoretical yield of H2CO3 by multiplying the moles of NaHCO3 used in the reaction by the molar mass of H2CO3:

Theoretical yield of H2CO3 = 0.04532 mol x 62.03 g/mol = 2.806 g

The percentage yield is calculated by dividing the actual yield (1.675 g) by the theoretical yield (2.806 g) and multiplying by 100:

Percentage yield of H2CO3 = (1.675 g / 2.806 g) x 100% = 59.7%

Therefore, the percentage yield of H2CO3 is 59.7%.