Given the reaction: p4(l) + 6 cl2(g) → 4 pcl3(l) if the percent yield is 82%, what mass of p4 is required to obtain 2.30 g pcl3 (cl2 in excess)?

The  mass  of P4   required to obtain 2.30  PCl3  0.694  g

  calculation

p4 + 6 Cl2→  4 pcl3

step 1:calculate the moles  of  PCl3  

moles=  mass/molar mass

from periodic table  the  molar mass   of PCl3  =  31 + (3 x35.5) = 137.5  g/mol

 moles = 2.30  / 137.5 =0.0167   moles

step 2: use the  mole  ratio to find the  moles of P4

from equation  above the  mole ratio  of p4 :PCl3   is  1:3  therefore  the  moles  of p4

    = 0.0167   x1/3  =0.0056 moles

step 3:  find the  mass of P4

mass = moles x   molar mass

from periodic table the   molar mass  of P4 =  31 x4= 124  g/mol

mass =  0.0056  x 124 = 0.694  g

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