Aqueous hydrochloric acid (HCl) reacts with solid sodium hydroxide (NaOH) to produce aqueous sodium chloride (NaCl) and liquid water (H2O). If 16.5 g of sodium chloride is produced from the reaction of 30.6g of hydrochloric acid and 19.5g of sodium hydroxide, calculate the percent yield of sodium chloride. Round your answer to 3 significant figures
The balanced equation for the reaction is:
HCl + NaOH -> NaCl + H2O
First, we need to determine the limiting reactant in the reaction. We can do this by comparing the moles of each reactant:
30.6 g HCl * (1 mol HCl / 36.4611 g HCl) = 0.8416 mol HCl
19.5 g NaOH * (1 mol NaOH / 39.9971 g NaOH) = 0.4876 mol NaOH
Since the molar ratio between HCl and NaOH is 1:1, we can see that HCl is in excess. Therefore, the limiting reactant is NaOH.
Next, we need to calculate the theoretical yield of NaCl.
0.4876 mol NaOH * (1 mol NaCl / 1 mol NaOH) * (58.44 g NaCl / 1 mol NaCl) = 28.487 g NaCl
Finally, we can calculate the percent yield.
Percent yield = (actual yield / theoretical yield) * 100
Percent yield = (16.5 g NaCl / 28.487 g NaCl) * 100 = 57.884%
Therefore, the percent yield of sodium chloride is 57.884%.