Answer:
The percent yield of this reaction is 96.8 %
Explanation:
Step 1: Data given
Mass of NaHCO3 = 10.0 grams
Mass of HCl = 10.0 grams
MAss of NaCl produced = 6.73 grams
Molar mass of NaHCO3 = 84.0 g/mol
Molar mass HCl = 36.46 g/mol
Molar mass NaCl = 58.44 g/mol
Step 2: The balanced equation
NaHCO3 (aq) + HCl (aq) → NaCl (aq) + CO2 (g) + H2O (l)
Step 3: Calculate moles
Moles = mass / molar mass
Moles NaHCO3 = 10.0 grams / 84.0 g/mol
Moles NaHCO3 = 0.119 moles
Moles HCl = 10.0 / 36.46 g/mol
Moles HCl = 0.274 moles
Step4: Calculate limiting reactant
For 1 mol NaHCO3 we need 1 mol HCl to produce 1 mol NaCl, 1 mol CO2 and 1 mol H2O
NaHCO3 is the limiting reactant. It will completely be consumed (0.119 moles). HCl is in excess. There will react 0.119 moles. There will remain 0.274 - 0.119 = 0.155 moles
Step 5: Calculate moles NaCl
For 1 mol NaHCO3 we need 1 mol HCl to produce 1 mol NaCl, 1 mol CO2 and 1 mol H2O
For 0.119 moles NaHCO3 we'll have 0.119 moles NaCl
Step 6: Calculate mass NaCl
Mass NaCl = moles NaCl * molar mass NaCl
Mass NaCl = 0.119 moles NaCl * 58.44 g/mol
Mass NaCl = 6.95 grams
Step 7: Calculate the percent yield
Percent yield =(actual mass / theoretical mass) * 100%
Percent yield = (6.73 grams / 6.95 grams) * 100%
PErcent yield = 96.8 %
The percent yield of this reaction is 96.8 %
