A reaction of 56.7 of Na and 39.3 g of Br2 yields 47.0 of NaBr. What is the percent yield?
This is a limiting reagent (LR) problem.
..................2Na + Br2 ==> 2NaBr
mols Na = 56.6/atomic mass Na = ?
mols Br2 = 39.3/molar mass Br2 = ?
Convert mols Na to mols NaBr produced knowing that 2 mol Na will produce 2 mols NaBr.
Do the same to convert mols Br2 to mols NaBr produced knowing that 1 mol Br2 will produce 2 mols NaBr
In LR problems, the SMALLEST number of mols NaBr produced is the correct shoice and the LR is the reactant that gives the smallest number of mols product formed.
Using the smallest number of mols NaBr produced, convert to grams NaBr. That is the theoretical yield (TY). The actual yield (AY) is given in the problem.
% yield = (AY/TY)*100 =
Post your work if you run into trouble.
I posted the above response.
To calculate the percent yield of a chemical reaction, you need to compare the actual yield (the amount of a product obtained from the reaction) with the theoretical yield (the maximum amount of product that could ideally be obtained).
In this case, you are given the masses of sodium (Na) and Bromine (Br2) that reacted, as well as the mass of Sodium Bromide (NaBr) obtained.
Step 1: Calculate the molar masses:
- The molar mass of Na is 22.99 g/mol.
- The molar mass of Br2 is 159.81 g/mol.
- The molar mass of NaBr is 102.89 g/mol.
Step 2: Convert the given masses to moles:
- Moles of Na = mass of Na / molar mass of Na
= 56.7 g / 22.99 g/mol = 2.47 mol
- Moles of Br2 = mass of Br2 / molar mass of Br2
= 39.3 g / 159.81 g/mol = 0.246 mol
Step 3: Use the balanced chemical equation to determine the stoichiometry between the reactants and products:
2 Na + Br2 -> 2 NaBr
According to the stoichiometry, for every 2 moles of Na, 2 moles of NaBr are produced.
Step 4: Calculate the theoretical yield of NaBr:
- Theoretical moles of NaBr = moles of Na x (2 moles of NaBr / 2 moles of Na)
= 2.47 mol x (2 mol NaBr / 2 mol Na) = 2.47 mol
- Theoretical mass of NaBr = moles of NaBr x molar mass of NaBr
= 2.47 mol x 102.89 g/mol = 254.19 g
Step 5: Calculate the percent yield:
- Percent yield = (actual yield / theoretical yield) x 100%
- Actual yield = 47.0 g
- Theoretical yield = 254.19 g
- Percent yield = (47.0 g / 254.19 g) x 100% = 18.48%
Therefore, the percent yield of the reaction is 18.48%.