Question

According to the following reaction, how many grams of iron(III) chloride will be formed upon the complete reaction of 27.1 grams of chlorine gas with excess iron?

The balanced eq is 2Fe+3Cl2=2FeCl3

Here is my worked out equation:

27.1g Cl*1 mol Cl/35.46 g Cl*2 mol FeCl3/3 mol Cl*162.2 g FeCl3/1 mol FeCl3

My answer is: 78.70 g FeCl3

What does it mean my excess iron? Do I have to incorporate that excess iron in the problem? Or did I solve it right?

Answer 1

No the excess is left over and does not participate.
27.1 g * 1 mol/35.5g = 0.763 mol Cl
for every mol Cl we use 1/3 mol Fe
so we actually used 1/3 * 0.763 = 0.254 mol Fe
0.254 mol * 56 g/mol = 14.2 grams Fe (the rest is excess left over)
so I get 27.1 + 14.2 = 41.3 grams of product

Answer 2

Thank you!