Percent yield in chemistry easy

Question

Aqueous hydrobromic acid HBr reacts with solid sodium hydroxide NaOH to produce aqueous sodium bromide NaBr and liquid water H2O. If 6.49g of sodium bromide is produced from the reaction of 13.8g of hydrobromic acid and 4.35g of sodium hydroxide, calculate the percent yield of sodium bromide.

Round your answer to 3 significant figures.

Answer 1

To calculate the percent yield of sodium bromide (NaBr), we need to compare the actual yield (the amount of NaBr produced in the reaction) to the theoretical yield (the amount of NaBr that would be produced if the reaction went to completion).

First, let's find the balanced chemical equation for the reaction:

HBr + NaOH -> NaBr + H2O

Next, we need to determine the limiting reactant, which is the reactant that will be completely consumed in the reaction and limits the amount of product formed.

Calculate the moles of each reactant:

Moles of HBr = mass of HBr (g) / molar mass of HBr (g/mol)

Moles of NaOH = mass of NaOH (g) / molar mass of NaOH (g/mol)

Molar masses:

HBr: 1 g/mol (hydrogen: 1 g/mol, bromine: 80 g/mol)

NaOH: 40 g/mol (sodium: 23 g/mol, oxygen: 16 g/mol, hydrogen: 1 g/mol)

Moles of HBr = 13.8 g / 81 g/mol ≈ 0.1704 mol

Moles of NaOH = 4.35 g / 40 g/mol ≈ 0.1088 mol

Determine the stoichiometric ratio between HBr and NaOH from the balanced equation:

1 mole of HBr reacts with 1 mole of NaOH.

The limiting reactant is the one with the smallest stoichiometric coefficient. In this case, NaOH has a smaller coefficient, indicating that it will be used up first.

Calculate the theoretical yield of NaBr based on the moles of NaOH:

The molar ratio between NaOH and NaBr is 1:1.

Theoretical moles of NaBr = moles of NaOH ≈ 0.1088 mol.

Calculate the theoretical mass of NaBr:

Theoretical mass of NaBr = moles of NaBr * molar mass of NaBr

Theoretical mass of NaBr = 0.1088 mol * 102.9 g/mol ≈ 11.19 g

Now, we can calculate the percent yield:

Percent yield = (Actual yield / Theoretical yield) * 100

Percent yield = (6.49 g / 11.19 g) * 100 ≈ 58.01%

The percent yield of sodium bromide is approximately 58.01%