If 16.9 g sodium carbonate is obtained from the thermal decomposition of 75.0 g of sodium hydrogen carbonate according to the unbalanced equation,

NaHCO3 �¨ Na2CO3 + H2O + CO2
What is the percent yield?

Here is a worked example of a stoichiometry problem INCLUDING percent yield.

http://www.jiskha.com/science/chemistry/stoichiometry.html

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To calculate the percent yield, we need to compare the amount of sodium carbonate obtained with the theoretical yield. The theoretical yield is the maximum amount of product that can be obtained, assuming a complete and efficient reaction.

Step 1: Calculate the molar mass of the compounds involved.
- Molar mass of NaHCO3 (sodium hydrogen carbonate) = 22.99 + 1.01 + 12.01 + (3 * 16.00) = 84.01 g/mol
- Molar mass of Na2CO3 (sodium carbonate) = (2 * 22.99) + 12.01 + (3 * 16.00) = 105.99 g/mol

Step 2: Determine the number of moles of sodium hydrogen carbonate used.
- Moles of NaHCO3 = mass (in grams) / molar mass
- Moles of NaHCO3 = 75.0 g / 84.01 g/mol = 0.892 mol

Step 3: Use stoichiometry to determine the theoretical yield of sodium carbonate.
From the balanced equation, we can see that one mole of NaHCO3 produces one mole of Na2CO3.
Thus, the theoretical yield of Na2CO3 is equal to the moles of NaHCO3 used.

Theoretical yield of Na2CO3 = 0.892 mol

Step 4: Calculate the actual yield.
The actual yield is given as 16.9 g.

Step 5: Calculate the percent yield.
Percent yield = (actual yield / theoretical yield) * 100
Percent yield = (16.9 g / (0.892 mol * 105.99 g/mol)) * 100

Using a calculator, we can find the percent yield:

Percent yield ≈ 18.3%

Therefore, the percent yield of sodium carbonate in this reaction is approximately 18.3%.