How many milliliters of 0.200 M FeCl3 are needed to react with an excess of Na2S to produce 0.345 g of Fe2S3 if the percent yield for the reaction is 65.0%? 3 Na2S(aq) + 2 FeCl3(aq) →→ Fe2S3(s) + 6 NaCl(aq)

25.53mL of 0.200 M FeCl₃ are needed to produce 0.345g of Fe₂S₃

Explanation:

Based on the reaction of the problem, 1 mole of Fe₂S₃ is produced from 2 moles of FeCl₃.

0.345g of Fe₂S₃ are (Molar mass: 207.9g/mol):

0.345g of Fe₂S₃ ₓ (1 mol / 207.9g) = 1.6595x10⁻³ moles Fe₂S₃

Moles of Fe needed to produce these moles of Fe₂S₃ are:

1.6595x10⁻³ moles Fe₂S₃ ₓ ( 2 moles FeCl₃ / 1 mole Fe₂S₃) =

3.3189x10⁻³ moles of FeCl₃

As the percent yield of the reaction is 65.0%, the moles of FeCl₃ you need to add are:

3.3189x10⁻³ moles of FeCl₃ ₓ (100.0% / 65.0%) = 5.106x10⁻³ moles of FeCl₃

A solution 0.200M contains 0.200 moles per L. Volume to obtain 5.106x10⁻³ moles is:

5.106x10⁻³ moles of FeCl₃ ₓ ( 1L / 0.200mol) = 0.02553L =

43.452

step-by-step explanation:

284 times .153 = 43.452