Question

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

Answer 1

I think this is the correct Answer

the balanced equation is

2 FeCl3 = 2 Fe + 3 Cl2

volume Fe = 78.4 mL / 7.87 g/mL = 9.96 g

moles Fe = 9.96 g / 55.847 g/mL=0.178

the ratio between FeCl3 and Fe is 2 : 2

moles FeCl3 = 0.178

mass FeCl3 = 0.178 x 162.206 g/mL=28.9 g

Not Completely sure though, But
Hope It Helps

Answer 2

Explanation:

The balanced reaction equation will be as follows.

        $2FeCl_{3} \rightarrow 2Fe + 3Cl_{2}$

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 = $\frac{mass}{volume}$

               7.87 g/ml = $\frac{mass}{78.4 ml}$

                     mass = 617 g

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

                    No. of moles = $\frac{mass}{\text{molar mass}}$

                                           = $\frac{617 g}{55.84 g/mol}$

                                           = 11.05 mol

Now, the ratio of $FeCl_{3}$ : Fe is 2:2.

Hence, number of moles of $FeCl_{3}$ will also be equal to 11.05 mol.

And, as  No. of moles = $\frac{mass}{\text{molar mass}}$

So, molar mass of $FeCl_{3}$ is 162.2 g/mol. Therefore, calculate the mass of $FeCl_{3}$ as follows.

                    No. of moles = $\frac{mass}{\text{molar mass}}$

                              11.05 mol = $\frac{mass}{162.2 g/mol}$

                             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.