Question

learn by doing
when 1.42 g of iron reacts with 1.90 g of chlorine, solid iron(iii) chloride and 7.15 kj of heat is produced, according to the equation below.
2 fe(s) + 3 cl₂(g) → 2 fecl₃(s)
how many moles of each reactant are available for the reaction?
mol fe
mol cl₂
which of the reactants is the limiting reactant? explain.

how many moles of fecl₃(s) will be produced in the reaction?
mol fecl₃
what is the enthalpy change, δh, for the reaction?
δh = kj

Explanation:

Step1: Calculate moles of Fe

Use formula \(n=\frac{m}{M}\), where \(m = 1.42g\) and \(M(Fe)=55.85g/mol\).
\(n(Fe)=\frac{1.42g}{55.85g/mol}\approx0.0254mol\)

Step2: Calculate moles of \(Cl_{2}\)

Use formula \(n=\frac{m}{M}\), where \(m = 1.90g\) and \(M(Cl_{2}) = 70.90g/mol\).
\(n(Cl_{2})=\frac{1.90g}{70.90g/mol}\approx0.0268mol\)

Step3: Determine limiting reactant

From equation \(2Fe + 3Cl_{2}
ightarrow2FeCl_{3}\), mole ratio \(Fe:Cl_{2}=2:3\).
For \(n(Fe) = 0.0254mol\), required \(n(Cl_{2})=\frac{3}{2}\times0.0254mol = 0.0381mol\).
For \(n(Cl_{2}) = 0.0268mol\), required \(n(Fe)=\frac{2}{3}\times0.0268mol\approx0.0179mol\).
Since available \(Cl_{2}(0.0268mol)<\) required \(Cl_{2}(0.0381mol)\), \(Cl_{2}\) is limiting.

Step4: Calculate moles of \(FeCl_{3}\)

From equation, \(n(FeCl_{3})=\frac{2}{3}n(Cl_{2})\).
\(n(FeCl_{3})=\frac{2}{3}\times0.0268mol\approx0.0179mol\)

Step5: Calculate \(\Delta H\)

From equation, \(2molFe\) (or \(3molCl_{2}\)) gives \(2molFeCl_{3}\).
\(0.0268molCl_{2}\) gives \(7.15kJ\).
For \(3molCl_{2}\), \(\Delta H=\frac{7.15kJ}{0.0268mol}\times3mol\approx - 800kJ\) (negative as heat is produced).

Answer:

mol Fe: \(0.0254\)
mol \(Cl_{2}\): \(0.0268\)
Limiting reactant: \(Cl_{2}\), as available \(Cl_{2}\) is less than required by \(Fe\).
mol \(FeCl_{3}\): \(0.0179\)
\(\Delta H\): \(- 800\)