When NaHCO3 completely decomposes, it can follow this balanced chemical equation:

2NaHCO3 → Na2CO3 + H2CO3

Determine the theoretical yields of each product using stoichiometry if the mass of the NaHCO3 sample is 3.24 grams. (Show work for both)
In an actual decomposition of NaHCO3, the mass of one of the products was measured to be 2.01 grams. Identify which product this could be and justify your reasoning.
Calculate the percent yield of the product identified in part B. (Show your work)

Question

When NaHCO3 completely decomposes, it can follow this balanced chemical equation:

2NaHCO3 → Na2CO3 + H2CO3

Determine the theoretical yields of each product using stoichiometry if the mass of the NaHCO3 sample is 3.24 grams. (Show work for both)
In an actual decomposition of NaHCO3, the mass of one of the products was measured to be 2.01 grams. Identify which product this could be and justify your reasoning.
Calculate the percent yield of the product identified in part B. (Show your work)

kekeman · Answer

@vocaloid

kekeman · Answer

@kekeman wrote:

@vocaloid

This is what I have came up with so far: PART A: The mass of NaHCO3 = 3.24 grams The molar mass of NaHCO3 = 84 g/mol Meaning the number of moles of NaHCO3 = 3.24/84 = 0.0386 mol The number of moles of Na2CO3 = 0.0386/2 = 0.0193 mol Na2CO3 = number of moles of Na2CO3 x molar mass of Na2CO3: 0.0193 x 106 ≈ 2.0458 g H2CO3 = number of moles of H2CO3 x molar mass of H2CO3: 0.0193 x 62 ≈ 1.1966 g PART B: The mass of one of the products was measured to be 2.01 grams and from the data above I have to say that the product has to be Na2CO3. I say this because 2.0458 g is the closest to 2.01 g. PART C: Percentage Yield of Na2CO3 = 2.01/ 2.0458 = 98%