Answer:
70.95%
Step-by-step explanation:
To calculate the percent yield of water in this reaction, we need to compare the actual amount of water produced to the theoretical amount of water that could be produced based on the amount of hydrobromic acid and sodium hydroxide used.
First, we need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed in the reaction, limiting the amount of product that can be formed.
To find the limiting reactant, we can use stoichiometry to calculate the amount of water that could be produced from each reactant, assuming they react completely. The balanced chemical equation for the reaction is:
HBr + NaOH → NaBr + H2O
From the equation, we can see that the mole ratio of HBr to H2O is 1:1, and the mole ratio of NaOH to H2O is 1:1. Therefore, the amount of water produced depends on the amount of HBr and NaOH present, and the reactant that produces less water is the limiting reactant.
Using the molar masses of the compounds, we can convert the masses of HBr and NaOH to moles:
moles of HBr = 17.0 g / 80.91 g/mol = 0.210 moles
moles of NaOH = 13.9 g / 40.00 g/mol = 0.348 moles
Based on the balanced chemical equation, the theoretical amount of water that could be produced from 0.210 moles of HBr is also 0.210 moles. The theoretical amount of water that could be produced from 0.348 moles of NaOH is also 0.348 moles.
However, since the amount of water produced is given as 2.69 g, we need to convert this to moles:
moles of H2O produced = 2.69 g / 18.02 g/mol = 0.149 moles
To calculate the percent yield of water, we can use the formula:
percent yield = (actual yield / theoretical yield) x 100%
where actual yield is the amount of water produced (0.149 moles) and theoretical yield is the amount of water that could be produced based on the limiting reactant.
Since the reactant that produces less water is the limiting reactant, we need to compare the theoretical yield of water from both reactants, and the lower value will be the theoretical yield based on the limiting reactant.
The theoretical yield of water from HBr is:
0.210 moles of HBr x (1 mole of H2O / 1 mole of HBr) = 0.210 moles of H2O
The theoretical yield of water from NaOH is:
0.348 moles of NaOH x (1 mole of H2O / 1 mole of NaOH) = 0.348 moles of H2O
Since the theoretical yield of water from HBr is lower, it is the limiting reactant. Therefore, the theoretical yield of water is 0.210 moles.
Now we can calculate the percent yield of water:
percent yield = (actual yield / theoretical yield) x 100%
percent yield = (0.149 moles / 0.210 moles) x 100%
percent yield = 70.95%
Therefore, the percent yield of water is 70.95%.