Enter a balanced equation for the reaction between aqueous lead(II)(II) nitrate and aqueous sodium chloride to form solid lead(I
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<u>Answer:</u> The percentage yield of HF is 73.36 %
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
....(1)
For calcium fluoride:
Given mass of calcium fluoride = 6.25 kg = 6250 g (Conversion factor: 1 kg = 1000 g)
Molar mass of calcium fluoride = 78.07 g/mol
Putting values in above equation, we get:
For the given chemical reaction:
By Stoichiometry of the reaction:
1 mole of calcium fluoride produces 2 moles of hydrofluoric acid
So, 80.05 moles of calcium fluoride will produce = of hydrofluoric acid
Now, calculating the theoretical yield of hydrofluoric acid using equation 1, we get:
Moles of of hydrofluoric acid = 160.1 moles
Molar mass of hydrofluoric acid = 20.01 g/mol
Putting values in equation 1, we get:
To calculate the percentage yield of hydrofluoric acid, we use the equation:
Experimental yield of hydrofluoric acid = 2.35 kg
Theoretical yield of hydrofluoric acid = 3.20 kg
Putting values in above equation, we get:
Hence, the percentage yield of HF is 73.36 %