Aqueous hydrobromic acid will react with solid sodium hydroxide to produce aqueous sodium bromide and liquid water . Suppose 4.05 g of hydrobromic acid is mixed with 0.86 g of sodium hydroxide. Calculate the maximum mass of water that could be produced by the chemical reaction. Be sure your answer has the correct number of significant digits.
HBr + NaO ==> NaBr + H2O
This is a limiting reagent (LR) problem and you know that because amounts are given for BOTH reactants.
mols HBr = grams/molar mass = ?
mols NaOH = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols HBr to mols H2O.
Do the same for mols NaOH to mols H2O.
It is likely that the two values for mols H2O will not be the same which means one of them is wrong; the correct value in LR problems is ALWAYS the smaller one and the reagent responsible for that value is the LR.
Now convert mols H2O to g. g = mols x molar mass.
To calculate the maximum mass of water produced, we need to determine the limiting reagent in the reaction. The limiting reagent is the reactant that will be completely consumed, thus determining the maximum amount of product that can be formed.
First, we need to calculate the number of moles for both hydrobromic acid and sodium hydroxide:
Molar mass of hydrobromic acid (HBr) = atomic mass of H + atomic mass of Br = 1.01 g/mol + 79.90 g/mol = 80.91 g/mol
Number of moles of HBr = mass of HBr / molar mass of HBr = 4.05 g / 80.91 g/mol ≈ 0.050 mol
Molar mass of sodium hydroxide (NaOH) = atomic mass of Na + atomic mass of O + atomic mass of H = 22.99 g/mol + 16.00 g/mol + 1.01 g/mol = 40.00 g/mol
Number of moles of NaOH = mass of NaOH / molar mass of NaOH = 0.86 g / 40.00 g/mol ≈ 0.0215 mol
Based on the balanced chemical equation, the mole ratio between HBr and water is 1:1. Therefore, the moles of water produced will be equal to the moles of HBr.
Number of moles of water = 0.050 mol
Finally, to calculate the mass of water produced, we need to multiply the number of moles of water by its molar mass:
Molar mass of water (H2O) = atomic mass of H x 2 + atomic mass of O = 1.01 g/mol x 2 + 16.00 g/mol = 18.02 g/mol
Mass of water = number of moles of water x molar mass of water = 0.050 mol x 18.02 g/mol ≈ 0.901 g
Therefore, the maximum mass of water that could be produced by the chemical reaction is approximately 0.901 grams.
To calculate the maximum mass of water produced, we need to determine which reactant is limiting, i.e., the one that is completely consumed in the reaction.
First, we need to find the number of moles of hydrobromic acid and sodium hydroxide using their respective molar masses:
Molar mass of hydrobromic acid (HBr):
H = 1.01 g/mol
Br = 79.90 g/mol
Molar mass of HBr = 1.01 g/mol + 79.90 g/mol = 80.91 g/mol
Number of moles of HBr = mass of HBr / molar mass of HBr
Number of moles of HBr = 4.05 g / 80.91 g/mol ≈ 0.05 mol
Molar mass of sodium hydroxide (NaOH):
Na = 22.99 g/mol
O = 16.00 g/mol
H = 1.01 g/mol
Molar mass of NaOH = 22.99 g/mol + 16.00 g/mol + 1.01 g/mol = 40.00 g/mol
Number of moles of NaOH = mass of NaOH / molar mass of NaOH
Number of moles of NaOH = 0.86 g / 40.00 g/mol ≈ 0.02 mol
Next, we need to balance the chemical equation to determine the stoichiometry of the reaction:
2HBr + 2NaOH → 2NaBr + 2H₂O
From the balanced equation, we can see that the molar ratio between HBr and H₂O is 2:2, which simplifies to 1:1. This means that for every 1 mole of HBr, 1 mole of H₂O is produced.
Since the stoichiometric ratio between HBr and H₂O is 1:1, the limiting reactant will be whichever one is present in the smaller amount. In this case, NaOH is the limiting reactant because it is present in a smaller quantity.
Now, we can calculate the maximum mass of water formed using the number of moles of NaOH:
Mass of water = moles of NaOH × molar mass of water
Mass of water = 0.02 mol × (2 × 1.01 g/mol + 16.00 g/mol) ≈ 0.86 g
Therefore, the maximum mass of water that could be produced by the chemical reaction is approximately 0.86 g.