Answer:
a) Theoretical Yield of HF = 5.64 grams
b) Percentage Yield = 39%
Explanation:
Reaction Given:
CaF2 + H2SO4 -> CaSO4 + 2HF
CaF2 = 11g
H2SO4 = Used in excess
HF = 2.2 g production = Actual Yield
So, Let's write down the molar masses:
Molar Mass of CaF2 = 78 g /mol
Molar Mass of HF = 20 g/mol
From the reaction, we can see the 1 mole of CaF2 gives the 2 moles of HF
i.e
a) Theoretical Yield of HF:
1 mole CaF2 = 2 moles HF
78 g CaF2 = 2 x 20 g of HF
78 g CaF2 = 40 g of HF
1 g CaF2 = 40g/78g of HF
And in the question it is given that chemist used 11 g of CaF2 so,
1 x 11 g of CaF2 = 11 x 40/78 g of HF
11 g of CaF2 = 440/78 g of HF
11 g of CaF2 = 5.64 g of HF
And this is the theoretical yield
Theoretical Yield of HF = 5.64 grams
b) Now, calculate the Percentage Yield of HF
Percentage Yield = Actual Yield /Theoretical Yield x 100
Percentage Yield = 2.2 g /5.64 g x 100
Percentage Yield = 39%
