A 3.24-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 2.19 grams.

Determine the mass of the H2CO3 produced.

Theoretical mass/yield of H2CO3 produced = 1.196 g

Actual mass/yield of  H2CO3 produced = 1.05 g

Explanation:

NaHCO3 sample was decomposed to  Na2CO3 + H2CO3. The balanced chemical equation for the reaction is represented below:

2NaHCO3 → Na2CO3 + H2CO3

The mass of NaHCO3 = 3.24 g

Molar mass of NaHCO3  = 23 + 1 + 12 + 48 = 84 g

From the chemical equation 2 moles of NaHCO3 was used. Therefore the mass will be 2 × 84 = 168 g

Molar mass of H2CO3 = 2 + 12 + 48 = 62 g

168 g of  NaHCO3 produces 62 g of H2CO3

3.24 g of NaHCO3 will produce ? g of H2CO3

cross multiply

theoretical mass of H2CO3 produced = (3.24 × 62)/168

theoretical mass of H2CO3 produced = 200.88/168

theoretical mass of H2CO3 produced = 1.1957142857 g

Theoretical mass of H2CO3 produced = 1.196 g

Actual mass of H2CO3

Actual mass of  H2CO3 produced = mass of NaHCO3 - mass of  Na2CO3

Actual mass of  H2CO3 produced = 3.24 - 2.19

Actual mass of  H2CO3 produced = 1.05 g

The question didn't indicate whether you want the actual mass or theoretical mass but i calculated for both mass.

it should be p1 v1 = p2 v2