81.9 % is the percent yield of reaction of zinc with iodine producing ZnI2.
Explanation:
Balanced chemical equation for the reaction:
Zn + I2 ⇒ ZnI2
Data given:
mass of sample zinc = 129 grams
mass of ZnI2 formed = 510.6 grams (actual yield)
number of moles = [tex]\frac{mass}{atomic mass of one mole}[/tex]
number of moles of zinc = [tex]\frac{129}{65.38}[/tex]
= 1.97 moles
from stoichiometry
1 mole of Zn reacts to form 1 mole of ZnI2
1.97 moles of Zn will form 1.97 moles of ZnI2
atomic mass of ZnI2 = 319.22 grams/mole
mass of ZnI2 formed = 628.86 grams (theoretical yield)
% yield = [tex]\frac{actual yield}{theoretical yield}[/tex] x100 equation 1
from the reaction,
Putting the values in equation 1
% yield = [tex]\frac{510.6}{628.86}[/tex] x 100
= 81.19 % is the % yield.
