Question

Equal weights of Zn metal and iodine are mixed together and is completely converted to. What fraction by weight of original Zn remains unreacted? (Zn = 65, I = 127)

Answer 1

The balanced chemical equation for the reaction between Zn and I2 is:

Zn + I2 → ZnI2

The molar mass of Zn is 65 g/mol and the molar mass of I2 is 254 g/mol. If equal weights of Zn and I2 are mixed together, then the number of moles of each reactant is:

Number of moles of Zn = Mass of Zn / Molar mass of Zn = (65 g) / (65 g/mol) = 1 mol
Number of moles of I2 = Mass of I2 / Molar mass of I2 = (65 g) / (254 g/mol) = 0.256 mol

According to the balanced chemical equation, 1 mol of Zn reacts with 1 mol of I2 to form 1 mol of ZnI2. Therefore, the number of moles of Zn that react is 0.256 mol, and the number of moles of Zn that remain unreacted is 1 - 0.256 = 0.744 mol.

The mass of the unreacted Zn is:

Mass of unreacted Zn = Number of moles of unreacted Zn x Molar mass of Zn
= (0.744 mol) x (65 g/mol)
= 48.36 g

Therefore, the fraction by weight of original Zn that remains unreacted is:

Fraction by weight of unreacted Zn = Mass of unreacted Zn / Total mass of Zn and I2
= (48.36 g) / (65 g + 65 g)
= 0.372 or 37.2%

Answer 2

Wasn't expecting that much but ok the answer I got was 0.74=74%

Answer 3

Yeah that's what I ended up with

Answer 4

Then yes, seems right.

The way Toga did it, there would be no excess Zn, as it would all be used up

Answer 5

oh ok