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 =

since electrons are negative the answer is neagtaive

nuclear fission produces elements that are heavier than helium.

the elements that are used in nuclear fission and their products are much heavier than helium.