In one process, 6.05 kg of caf2 is treated with an excess of h2so4 and yields 2.85 kg of hf. calculate the percent yield of hf.

 The %  yield  of Hf  is  91. 8%

calculation

% yield =  actual mass/ theoretical mass  x100
actual mass = 2.85 Kg

calculate the theoretical  mass
by first  write the equation for reaction

CaF2  +H2SO4  → 2HF  + CaSO4
find  the  moles  of CaF2  used
moles = mass/molar mass
mass = 6.05Kg   = 6.05 x1000 =6050 Kg
molar mass  of CaF2  =  40 + (19 x2) = 78 g/mol

moles= 6050/ 78 = 77.6 moles

by  use  of mole ratio between  CaF2  to HF  which  is  1:2 the  moles of HF is therefore = 77.6 x2 =155.2  moles of Hf

find the  theoretical mass  of HF = moles x  molar mass ( 1 +19=20g/mol)

=   155.2  moles x 20 g/mol = 3104 grams  = 3104 /1000 = 3.104 Kg

The % yield  is therefore  = 2.85 Kg/ 3.104 Kg x100 =  91.8%

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