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.
A. Determine the mass of the H2CO3 produced.
B. Calculate the percentage yield of H2CO3 for the reaction. Show your work or describe the calculation process in detail.

A. 1.67g H2CO3

B.117.6%

Explanation:

To solve this question we need to find the moles of Na2CO3 produced = Moles of H2CO3 produced. With the moles and molar mass of H2CO3 -62.03g/mol- we can find its mass produced -Actual yield-.

Theoretical mass must be obtained in the same way but from the mass of NaHCO3 produced.

Percent yield is 100 times the ratio of actual yield and theoretical mass:

Moles Na2CO3 -Molar mass: 105.99g/mol-

2.86g * (1mol/105.99g) = 0.02698 moles Na2CO3 = Moles H2CO3

Mass H2CO3 = Actual yield:

0.02698 moles * (62.03g/mol) = 1.67g H2CO3

Moles NaHCO3 -Molar mass: 84.007g/mol-

3.81g * (1mol/84.007g) = 0.04535 moles NaHCO3

Moles H2CO3:

0.04535 moles NaHCO3 * (1mol H2CO3/2molNaHCO3) = 0.02268moles H2CO3

Mass H2CO3 = Theoretical yield:

0.02268 moles * (62.03g/mol) = 1.41g H2CO3

Percent yield:

1.67g / 1.42g * 100 = 117.6%

the answer is b. dilution