For the balanced equation shown below, if the reaction of 20.7 grams of caco3 produces 6.81 grams of cao, what is the percent yield? > cao+co2

The  %  yield  is  calculated  as  using  the  below  formula

actual  yield/theoretical   yield  x100
actual  yield =  6.81

   the  theoretical  yield is   calculated  as  follows

find  the  moles  of  CaCO3
moles  =  mass/molar  mass
=  20.7g/100g/mol  =  0.207  moles
CaCO3  = CaO  + CO2
by   use  of  mole  ratio  between   CaCO3  to  CaO  which  is  1:1  the   theoretical moles  of  CaO  is  therefore   =  0.207
the  theoretical  mass  is therefore  =  theoretical  moles  x  molar  mass of CaO
=  0.207  x  56 =  11.  59  g

%  yield  is therefore=  6.81/11.59  x100=  58.76%

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