How many grams of iron (iii) chloride must decompose to produce 78.4 milliliters of iron metal, if the density of iron is 7.87 g/ml? show all steps of your calculation as well as the final answer. fecl3 → fe + cl2

Explanation:

The balanced reaction equation will be as follows.

As we known that density is mass divided by volume of the substance. And, it is given that density is 7.87 g/ml and volume is 78.4 ml.

Hence, calculate the mass as follows.

                  Density =

               7.87 g/ml =

                     mass = 617 g

Therefore, number of moles of iron present in 617 g will be as follows.

                    No. of moles =

                                           =

                                           = 11.05 mol

Now, the ratio of : Fe is 2:2.

Hence, number of moles of will also be equal to 11.05 mol.

And, as  No. of moles =

So, molar mass of is 162.2 g/mol. Therefore, calculate the mass of as follows.

                    No. of moles =

                              11.05 mol =

                             mass = 1792.31 g

Thus, we can conclude that 1792.31 g of iron (III) chloride must decompose to produce 78.4 milliliters of iron metal, if the density of iron is 7.87 g/mL.