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Aqueous hydrochloric acid reacts with solid sodium hydroxide to produce aqueous sodium chloride and liquid water . What is the theoretical yield of water formed from the reaction of 0.73 g of hydrochloric acid and 0.66 g of sodium hydroxide?rnrnround your answer to 2 significant figures.
Accepted Answer
Here is the step-by-step solution to this problem:
Step 1: Balance the chemical equation
HCl(aq) + NaOH(aq) --→ NaCl(aq) + H2O(l)
Step 2: Convert the masses of the reactants to moles
moles HCl = 0.73 g HCl * (1 mol HCl / 36.46 g HCl) = 0.02 mol HCl
moles NaOH = 0.66 g NaOH * (1 mol NaOH / 40.00 g NaOH) = 0.0165 mol NaOH
Step 3: Determine the limiting reactant
To do this, we will compare the number of moles of each reactant to the stoichiometric coefficients in the balanced chemical equation.
For HCl: 0.02 mol HCl / 1 mol HCl = 0.02
For NaOH: 0.0165 mol NaOH / 1 mol NaOH = 0.0165
Because NaOH has the smaller value, it is the limiting reactant.
Step 4: Calculate the theoretical yield of water
moles H2O = 0.0165 mol NaOH * (1 mol H2O / 1 mol NaOH) = 0.0165 mol H2O
mass H2O = 0.0165 mol H2O * (18.02 g H2O / 1 mol H2O) = 0.297 g H2O
Therefore, the theoretical yield of water formed from the reaction of 0.73 g of hydrochloric acid and 0.66 g of sodium hydroxide is 0.30 g.
Rounded to 2 significant figures, the theoretical yield of water is 0.30 g.
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