During a laboratory experiment, a 3.81-gram sample of nahco3 was thermally decomposed. in this experiment, carbon dioxide and water vapors escape and are combined to form carbonic acid. after decomposition, the sample weighed 2.86 grams. calculate the percentage yield of carbonic acid for the reaction. describe the calculation process in detail. nahco3 → na2co3 + h2co3

The % yield is 24.4%. See below for details.

Explanation:

First balance the chemical formula:

2NaHCO3 --> Na2CO3 + H2CO3

Next calculate the moles of NaHCO3 at the beginning and end by dividing by molar mass:

Initial: 3.81 / 84 = 0.045 mol

Final: 2.86 / 84 = 0.034 mol

Change (actual amount reacted): 0.045 - 0.034 = 0.011 mol

You can see based on the original chemical formula that for every 2 moles of NaHCO3 that react, only 1 mole of H2CO3 is produced. Therefore, you will need to divide each NaHCO3 value by 2 to find the corresponding value of H2CO3. In this case we are most interested in the initial amount of NaHCO3 (since this represents the maximum possible amount of reactant, and therefore product, we can have - the theoretical amount of product), and the change in NaHCO3 (since this is the amount of reactant that actually reacted):

Theoretical H2CO3: 0.045 / 2 = 0.0225 mol

Actual H2CO3: 0.011 / 2 = 0.0055 mol

The percent yield is calculated from these two values (you could convert back to mass if you wanted but it wouldn't change the answer):

% yield = actual / theoretical

% yield = 0.0055/0.0225

% yield = 24.4%

the answer is b & c i believe : )

your answer is a. the enthalpy of formation for a pure element in its standard state is always positive.