Yield = practical/theoretical
Given:
Practical mass: Pm{ClF3} = 4.6g
Reactant mass: m{F2} = 5.3g
The equation: 3 F2 + Cl2 = 2 ClF3
Lookup the Molecular Weights:
w{F2} = 37.99680650 ± 0.00000002 g/mol
w{ClF3} = 92.4484 ± 0.0001 g/mol
So theoretically, the mass of chlorine triflouride is:
Tm{ClF3} = (2/3) m{F2}w{ClF3}/w{F2}
Therefore:
Yield = Pm{ClF3}/Tm{ClF3}
= (3/2) Pm{ClF3} w{F2} /(m{F2} w{ClF3})
= (3/2)*4.6*37.99680650/(5.3*92.4484)
= 53.5%
The reaction of 5.3 grams of fluorine with
excess chlorine produced 4.6 grams of ClF3. What percent yield of ClF3 was obtained?
Answer in units of %
1 answer
1 answer