Answer
The percent yield of HF is equal to 92.9% for this reaction.
Work Step by Step
1. Calculate the molar masses:
$ CaF_2 $ : ( 40.08 $\times$ 1 )+ ( 19.00 $\times$ 2 )= 78.08 g/mol
$$ \frac{1 \space mole \space CaF_2 }{ 78.08 \space g \space CaF_2 } \space and \space \frac{ 78.08 \space g \space CaF_2 }{1 \space mole \space CaF_2 }$$
$ HF $ : ( 1.008 $\times$ 1 )+ ( 19.00 $\times$ 1 )= 20.01 g/mol
$$ \frac{1 \space mole \space HF }{ 20.01 \space g \space HF } \space and \space \frac{ 20.01 \space g \space HF }{1 \space mole \space HF }$$
2. Calculate the theoretical yield of HF:
$$ 6.00 \space kg \space CaF_2 \times \frac{1 \space mole \space CaF_2 }{ 78.08 \space g \space CaF_2 } \times \frac{ 2 \space moles \space HF }{ 1 \space mole \space CaF_2 } \times \frac{ 20.01 \space g \space HF }{1 \space mole \space HF } = 3.08 \space kg \space HF $$
3. Calculating the percent yield: $$\frac{ 2.86 \space kg \space HF }{ 3.08 \space kg \space HF } \times 100\% = 92.9 \%$$
