Zn + I2 ------------> ZnI2
1 moles of Zn react with I2 to gives 1 mole of ZnI2
0.5moles of Zn react with I2 to gives = 1*0.5/1 = 0.5 moles of
ZnI2
actual yield yield of ZnI2Â Â = no of moles * molar
mass of ZnI2
                                   Â
= 0.5*319.22
                                     Â
= 159.61g
percent yield = actual yield *100/theoretical yield
                 Â
= 159.61*100/515.6Â Â = 30.96%
>>>>>answer
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