Aqueous hydrochloride acid (HCl) reacts with sodium hydroxide (NaOH) to produce aqueous sodium chloride (NaCl) and Louis water (H2O). If 5.39 g of sodium is produced of 17.5 g of hydrochloric acid and 9.22 g of sodium hydroxide, calculate the percent yield of sodium chloride. Round your answer to 3 significant figures
To determine the percent yield of sodium chloride, we need to calculate the theoretical yield of sodium chloride first.
1. Start by finding the moles of hydrochloric acid (HCl) using its molar mass.
Molar mass of HCl = 1.007 g/mol (H) + 35.453 g/mol (Cl) = 36.460 g/mol
Moles of HCl = 17.5 g / 36.460 g/mol = 0.480 mol
2. Next, find the moles of sodium hydroxide (NaOH) using its molar mass.
Molar mass of NaOH = 22.990 g/mol (Na) + 15.999 g/mol (O) + 1.007 g/mol (H) = 39.996 g/mol
Moles of NaOH = 9.22 g / 39.996 g/mol = 0.231 mol
3. The balanced equation for the reaction is:
HCl + NaOH -> NaCl + H2O
From the equation, we see that the mole ratio between HCl and NaCl is 1:1. Therefore, the moles of NaCl produced will be the same as the moles of HCl used, which is 0.480 mol.
4. Find the molar mass of sodium chloride (NaCl).
Molar mass of NaCl = 22.990 g/mol (Na) + 35.453 g/mol (Cl) = 58.443 g/mol
5. Calculate the theoretical yield of sodium chloride.
Theoretical yield (NaCl) = 0.480 mol x 58.443 g/mol = 28.01 g
6. Finally, calculate the percent yield.
Percent Yield = (Actual Yield / Theoretical Yield) x 100%
Actual Yield = 5.39 g (given)
Percent Yield = (5.39 g / 28.01 g) x 100%
Percent Yield = 19.245%
Rounded to three significant figures, the percent yield of sodium chloride is 19.2%.