how many milliliters of 0.200 m fecl3 are needed to react with an excess of na2s to produce 2.75 g of fe2s3 if the percent yield for the reaction is 65.0%?

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