Question

Consider the following reaction:
iron (s) + chlorine (g) à iron (III) chloride
a. Write the balanced chemical equation.
b. 25.0 g of iron reacts with excess chlorine gas.
a. Calculate the moles of iron reactant.
b. Calculate the moles of iron (III) chloride.
c. Calculate the mass of iron (III) chloride.

Answer 1

Answer:

1) a. 2Fe(s) + 3Cl₂(g) → 2FeCl₃(s).

2) a. 0.4477 mol.

b. 0.4477 mol.

c. 72.61 g.

Explanation:

1) Consider the following reaction:  iron (s) + chlorine (g) à iron (III) chloride

a. Write the balanced chemical equation.

• The balanced equation should apply the law of conversation of mass that the no. of different atoms is equal in both sides of the reaction (reactants and products sides).

So, the balanced chemical equation is:

2Fe(s) + 3Cl₂(g) → 2FeCl₃(s).

It is clear that 2 mol of Fe(s) react with 3 mol of Cl₂(g) to produce 2 mol of FeCl₃(s).

2) 25.0 g of iron reacts with excess chlorine gas.  

a. Calculate the moles of iron reactant.

• The no. of moles of Fe (n) can be calculated using the relation:

n = mass/molar mass = (25.0 g)/(55.845 g/mol) = 0.4477 mol.

b. Calculate the moles of iron (III) chloride.

Using cross multiplication:

2 mol of Fe produce → 2 mol of FeCl₃, from stichiometry.

∴ 0.4477 mol of Fe produce → 0.4477 mol of FeCl₃.

∴ The no. of moles of iron (III) chloride produced is 0.4477 mol.

c. Calculate the mass of iron (III) chloride.

• We can calculate the mass of iron (III) chloride produced using the relation:

mass of  iron (III) chloride = (no. of moles)*(molar mass) = (0.4477 mol)*(162.2 g/mol) = 72.61 g.