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Aqueous hydrobromic acid reacts with solid sodium hydroxide to produce aqueous sodium bromide and liquid water . What is the theoretical yield of water formed from the reaction of 4.9 g of hydrobromic acid and 1.9 g of sodium hydroxide?rnrnround your answer to 2 significant figures.
Accepted Answer
To determine the theoretical yield of water formed, we need to first balance the chemical equation:
2HBr(aq) + 2NaOH(s) → 2NaBr(aq) + 2H2O(l)
From the balanced equation, we can see that 2 moles of HBr react with 2 moles of NaOH to produce 2 moles of H2O.
1. Convert the mass of HBr to moles:
moles HBr = 4.9 g / 80.91 g/mol = 0.0606 mol
2. Convert the mass of NaOH to moles:
moles NaOH = 1.9 g / 40 g/mol = 0.0475 mol
3. Determine the limiting reactant:
To do this, we compare the mole ratio of HBr to NaOH to the mole ratio in the balanced equation. The reactant that produces the smaller mole ratio is the limiting reactant.
From the balanced equation, we know that 2 moles of HBr react with 2 moles of NaOH.
Actual mole ratio = moles HBr / moles NaOH = 0.0606 mol / 0.0475 mol = 1.28
Since 1.28 is closer to 1 than 2, HBr is the limiting reactant.
4. Calculate the theoretical yield of water:
From the balanced equation, we know that 2 moles of HBr produce 2 moles of H2O.
Using the mole ratio, we can calculate the moles of H2O produced:
moles H2O = 0.0606 mol HBr * (2 mol H2O / 2 mol HBr) = 0.0606 mol H2O
5. Convert the moles of H2O to grams:
grams H2O = 0.0606 mol H2O * 18 g/mol = 1.09 g
Therefore, the theoretical yield of water formed from the reaction of 4.9 g of hydrobromic acid and 1.9 g of sodium hydroxide is 1.09 g.
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