Answer:
We need 203 mL of FeCl3
Explanation:
Step 1: Data given
molarity FeCl3 = 0.200 M
Mass of Fe2S3 = 2.75 grams
Percent yield = 65.0 %
Step 2: The balanced equation
3 Na₂S(aq) + 2 FeCl₃(aq) → Fe₂S₃(s) + 6 NaCl(aq)
Step 3: Calculate moles Fe2S3
Moles Fe2S3 = mass Fe2S3 / molar mass Fe2S3
Moles Fe2S3 = 2.75 grams / 207.9 g/mol
Moles Fe2S3 = 0.0132 moles
Step 4: Calculate theoretical yield
65.0 % = 0.65 = actual yield/ theoretical yield
theoretical yield = actual yield / 0.65
theoretical yield = 0.0132 moles /0.65
theoretical yield = 0.0203 moles Fe2S3
Step 5: Calculate moles FeCl3
For 1 mol Fe2S3 and 6 mol NaCl we need 3 moles Na2S and 2 moles FeCl3
For 0.0203 moles Fe2S3 we need 2*0.0203 = 0.0406 moles FeCl3
Step 6: Calculate volume
Molarity = moles / volume
Volume = moles / molarity
Volume = 0.0406 moles / 0.200 M
Volume = 0.203 L = 203 mL
We need 203 mL of FeCl3
