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Aqueous hydrobromic acid reacts with solid sodium hydroxide to produce aqueous sodium bromide and liquid water . if 12.6 g of water is produced from the reaction of 72.8 g of hydrobromic acid and 71.3 g of sodium hydroxide, calculate the percent yield of water? round your answer to 3 significant figures. What is the answer?
Accepted Answer
Here's how to calculate the percent yield of water:
1. Write the balanced chemical equation:
HBr(aq) + NaOH(aq) → NaBr(aq) + H₂O(l)
2. Calculate the theoretical yield of water:
Convert the masses of HBr and NaOH to moles using their molar masses:
Moles of HBr = 72.8 g / 80.91 g/mol = 0.900 mol
Moles of NaOH = 71.3 g / 40.00 g/mol = 1.78 mol
Determine the limiting reactant. Since the stoichiometric ratio is 1:1, the limiting reactant is HBr because it has fewer moles.
Calculate the moles of water produced from the limiting reactant (HBr):
Moles of H₂O = 0.900 mol HBr
(1 mol H₂O / 1 mol HBr) = 0.900 mol H₂O
Convert the moles of water to grams using its molar mass:
Theoretical yield of H₂O = 0.900 mol
18.02 g/mol = 16.2 g
3. Calculate the percent yield:
Percent yield = (Actual yield / Theoretical yield)
100%
Percent yield = (12.6 g / 16.2 g)
100% = 77.8%
Therefore, the percent yield of water is 77.8%.
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