Q4.104 P - Chemistry: Molecular Nature Of Matter And Change

Question

Iron reacts rapidly with chlorine gas to form a reddish brown, ionic compound (A), which contains iron in the higher of its two common oxidation states. Strong heating decomposes compound A to compound B, another ionic compound, which contains iron in the lower of its two oxidation states. When compound A is formed by the reaction of 50.6 g of Fe and 83.8 g of Cl₂ and then heated, how much compound B forms?

Step-by-Step Solution

Answer

You need to calculate when compound A is formed by the reaction of 50.6 g of Fe and 83.8 g of Cl₂ and then heated, how much compound B forms.

Step 1: Balanced equation

Iron reacts with chlorine and forms iron chloride (compound A). The balanced equation is

$3Fe\left(s\right)+C{I}_2\left(g\right)\to 2FeC{I}_3\left(s\right)$

Step 2: Calculation of moles of Fe

According to the question,

Mass of Fe = 50.6g

Molecular mass of Fe = 55.85g/mol

Again you know,

$moles=\frac{mass\left(given\right)}{mass\left(molar\right)}$

Moles of Fe

$$\begin{gathered} =\frac{50.6}{55.85}(mol \\ =0.906(mol) \end{gathered}$$

Hence, moles of Fe are 0.906mol.

Step 3: Calculation of moles of FeCl 3 when Fe is limiting reagent

According to the balanced equation

$3Fe\left(s\right)+C{I}_2\left(g\right)\to 2FeC{I}_3\left(s\right)$

3 moles of Fe produces 2 moles FeCl₃

Now, moles of FeCl₃

$$\begin{gathered} =0.906\times \frac{2}{2}\left(mol\right) \\ =0.906\left(mol\right) \end{gathered}$$

Hence, moles of FeCl₃ when Fe is limiting reagent are 0.906mol.

Step 4: Calculation of moles of Cl 2

According to the question,

Mass of Cl₂ = 83.8g

Molecular mass of Cl₂ = 70g/mol

Again you know,

$moles=\frac{mass\left(given\right)}{mass\left(molar\right)}$

Moles of Cl₂

$$\begin{gathered} =\frac{83.8}{70}\left(mol\right) \\ =1.197\left(mol\right) \end{gathered}$$

Hence, moles of Cl₂ are 1.197mol.

Step 5: Calculation of moles of FeCl 3 when Cl 2 is limiting reagent

According to the balanced equation

$3Fe\left(s\right)+C{I}_2\left(g\right)\to 2FeC{I}_3\left(s\right)$

1 mole of Cl₂ produces 2 moles FeCl₃

Now, moles of FeCl₃

$$\begin{gathered} =1.197\times \frac{2}{3}\left(mol\right) \\ =0.798\left(mol\right) \end{gathered}$$

Hence, moles of FeCl₃ when Cl₂ is limiting reagent are 0.798mol.

Step 6: Conclusion

Cl₂ is the limiting reagent in this reaction as moles of FeCl₃ is less in terms of Cl₂.

Step 7: Calculation of moles of compound B

FeCl₃ dissociates to produce FeCl₂ (compound B) and Cl₂. The balanced equation is

$2FeC{I}_3\left(A\right)\to 2FeC{I}_2\left(B\right)+C{I}_2$

Moles of FeCl₃ = 0.798moles

According to the equation,

2 moles of FeCl₃ produces 2 moles of FeCl₂

Now moles of FeCl₂

$$\begin{gathered} =0.798\times \frac{2}{2}\left(mol\right) \\ =0.798\left(mol\right) \end{gathered}$$

Step 8: Calculation of mass of compound B

Molecular mass of FeCl₂ = 125.85g/mol

Again you know,

mass = moles x mass(molar)

Mass of FeCl₂

$$\begin{gathered} =0.798\times 125.85\left(g\right) \\ =100.4283\left(g\right) \end{gathered}$$

Hence, mass of FeCl₂ is 100.4283.

Step 9: Conclusion

When compound A is formed by the reaction of 50.6 g of Fe and 83.8 g of Cl₂ and then heated, the mass of compound B is 100.4283g.