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

h let me think

since the atomic number is the number of protons in the nucleus, as we move to the left on the periodic table, the electrostatic pull between the nucleus and the valence electron shell decreases. as we move down a group (vertical column), the valence shell energy level gets further and further from the nucleus.