equal weight of zn metal & iodine are mixed together & iodine is completely converted to zni2.what fraction of weight of the original zinc remains unreated

Gaurav
askIITians Faculty 164 Points
9 years ago
The reaction can be represented by the equation
Zn(s) + I2(s) → ZnI2(s)
Lets take a sample of 100 g each

100 g Zn / 63.4 g/mol = 1.58 moles
100 g I2 / 253.8 g/mol = 0.394 mol I2

Since these two react in a 1:1 ratio, you will have used 0.394 moles of Zn to react with the 0.394 moles of I2. That would leave 1.58 - 0.394 = 1.186 moles of Zn unreacted.

So, the fraction of the original zinc remaining would be 1.186 / 1.58 = .75
Himanshu gupta
30 Points
7 years ago
student-name Abhishek Singh asked in Chemistryequal weight of Zn metal and iodine are miced together and the iodine is completely converted to ZnI(2). what fraction of weight of the original Zn remains unreacted. 1 Follow 0student-name Kumar Sarang answered this1279 helpful votes in Chemistry, Class XReaction involved-Zn(s) + I2(s) → ZnI2(s)Considering the 100 g of Zn react with 100 g of I2Number of moles of Zn= Mass of Zn / molar mass of Zn= 100 g / 65 g/mol= 1.5384 molesNumber of moles of I2= Mass of I2/molar mass of I2= 100 g /253.8 g/mol= 0.394 molesThus, 1.5384 moles of Zn react with 0.394 moles of I2 and thus 1.186 moles of Zn unreacted.Fraction of weight of the original Zn remains unreacted= 1.18/1.54= 0.7662