PLZZZ HELP! 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. Calculate the percentage yield of H2CO3 for the reaction. Show your work or describe the process in detail

Mass of H2CO3 produced = 1.05 g

Percentage yield of H2CO3 = 87.5%

Explanation:

Equation of the reaction is given below:

2NaHCO3 -> Na2CO3 + H2CO3

From the equation of reaction , 2 moles of NaHCO3 decomposes to produce 1 mole of H2CO3

Molar mass of NaHCO3 = 23 + 1 + 12 + 16 × 3 =  84 g

Molar mass of H2CO3 = 1 × 2 + 12 + 16 × 3 = 62 g

2 moles of NaHCO3 = 2 × 84 g = 168 g

Therefore,  168 g of NaHCO3 will produce 62 g of H2CO3

3.24 g of NaHCO3 will produce 62/168 × 3.24 g of H2CO3  = 1.20 g of H2CO3

From the value given in the question, 3.24 g of NaHCO3 decomposes completely to give Na2CO3 and H2CO3.

2.19 g of Na2CO3 are produced.

From the law of conservation of mass, mass of H2CO3 produced = 3.24 g - 2.19 g

Mass of H2CO3 produced = 1.05 g

Percentage yield = actual yield/theoretical yield × 100%

Actual yield = 1.05 g; Theoretical yield = 1.20 g

Percentage yield  = 1.05/1.20 × 100%

Percentage yield  = 87.5%