How many milliliters of 0.200 m fecl3 are needed to react with an excess of na2s to produce 1.38 g of fe2s3 if the percent yield for the reaction is 65.0%? 3 na2s(aq) + 2 fecl3(aq) → fe2s3(s) + 6 nacl(aq)?

100 ml


Explanation:

1) Convert 1.38 g of Fe₂S₃ into number of moles, n

i) Formula: n = mass in grass / molar mass

ii) molar mass of Fe₂S₃ =2 x 55.8 g/mol + 3 x 32.1 g/mol = 207.9 g/mol


iii) n = 1.38 g / 207.9 g/mol = 0.00664 moles of Fe₂S₃


2) Use the percent yield to calculate the theoretical amount:


65% = 0.65 = actual yield/ theoretical yield =>

theoretical yield = actual yield / 0.65 = 0.00664 moles / 0.65 = 0.010 mol Fe₂S₃

3) Chemical equation:


3 Na₂S(aq) + 2 FeCl₃(aq) → Fe₂S₃(s) + 6 NaCl(aq)

4) Stoichiometrical mole ratios:


3 mol Na₂S : 2 mol FeCl₃ : 1 mol Fe₂S₃ : 6 mol NaCl

5) Proportionality:

2moles FeCl₃ / 1 mol Fe₂S₃ = x / 0.010 mol Fe₂S₃

=> x = 0.020 mol FeCl₃

6) convert 0.020 mol to volume


i) Molarity formula: M = n / V


ii) V = n / M = 0.020 mol / 0.2 M = 0.1 liter = 100 ml

The correct answer is b