Question

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)?

Answer 1

Answer: The volume of ferric chloride needed is 102 mL

Explanation:

To calculate the theoretical yield of ferric sulfide, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of ferric sulfide = 1.38 g

Percentage yield of ferric sulfide = 65.0 %

Putting values in above equation, we get:

$65.0=\frac{1.38g}{\text{Theoretical yield of ferric sulfide}}\times 100\\\\\text{Theoretical yield of ferric sulfide}=\frac{1.38}{65}\times 100=2.12g$

• To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of ferric sulfide = 2.12 g

Molar mass of ferric sulfide = 208 g/mol

Putting values in above equation, we get:

$\text{Moles of ferric sulfide}=\frac{2.12g}{208g/mol}=0.0102mol$

The given chemical equation follows:

$3Na_2S(aq.)+2FeCl_3(aq.)\rightarrow Fe_2S_3(s)+6NaCl(aq.)$

By Stoichiometry of the reaction:

1 mole of ferric sulfide is produced from 2 moles of ferric chloride

So, 0.0102 moles of ferric sulfide will be produced from $\frac{2}{1}\times 0.0102=0.0204mol$ of ferric chloride

• To calculate the volume for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$

Molarity of ferric chloride = 0.200 M

Moles of ferric chloride = 0.0204 moles

Putting values in above equation, we get:

$0.200M=\frac{0.0204\times 1000}{\text{Volume of ferric chloride}}\\\\\text{Volume of ferric chloride}=\frac{0.0204\times 1000}{0.200}=102mL$

Hence, the volume of ferric chloride needed is 102 mL

Answer 2

Answer: 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