Question

Consider the reaction: 4Fe(s) + 3O2(g) à2Fe2O3(s) DHo= -1652 kJHow much heat is involved in the process if 25.6 g of Fe(s) reacts with 13.5 g of O2(g)

Answer 1

Reaction is

4 Fe(s) + 3 O2(g) .......................> 2 Fe2O3(s)

4 mole Fe react with 3 mole O2 to form 2 mole Fe2O3.

25.6 g of Fe(s) = mass / molar mass = 25.6 g / 55.85 g / mole = 0.458 mole.

13.5 g of O2(g) = 13.5 g / 32 g / mole = 0.4219 mole.

Fe is the limiting reactant.

for reaction of 4 mole Fe, heat released = 1652 kJ

thus

heat involved in this process = 1652 KJ * 0.458 mole / 4 mole = 189.1 KJ (answer)

Answer 2

To find out how much heat is involved in the given reaction when 25.6 g of Fe(s) reacts with 13.5 g of O2(g), we first need to determine which reactant is limiting and then calculate the amount of heat released based on the stoichiometry of the balanced reaction.

1. Calculate the moles of each reactant:

Molar mass of Fe (iron) = 55.845 g/mol Molar mass of O2 (oxygen) = 32.00 g/mol

Moles of Fe = 25.6 g / 55.845 g/mol Moles of Fe ≈ 0.4583 mol

Moles of O2 = 13.5 g / 32.00 g/mol Moles of O2 ≈ 0.4219 mol

2. Determine the limiting reactant:

To find the limiting reactant, we need to compare the stoichiometric ratio of the reactants in the balanced equation with the actual ratio of moles present.

The balanced equation is: 4Fe(s) + 3O2(g) → 2Fe2O3(s)

From the balanced equation, we see that the ratio of moles between Fe and O2 is 4:3. Therefore, we can use the moles of Fe and O2 to determine which reactant is limiting.

Ratio of moles of Fe to O2 = 0.4583 mol / 0.4219 mol ≈ 1.087

Since the ratio is approximately 1.087, it means that there are more moles of Fe than O2. Thus, O2 is the limiting reactant.

3. Calculate the amount of heat involved using the molar ratio and enthalpy change:

The balanced equation tells us that 4 moles of Fe react with 3 moles of O2 to produce 2 moles of Fe2O3. Since O2 is the limiting reactant, we will use its moles to calculate the heat involved.

According to the balanced equation: 4 moles of Fe react with 3 moles of O2, and the enthalpy change is -1652 kJ.

Moles of O2 reacting = 0.4219 mol (since O2 is limiting)

Now, we can use stoichiometry to calculate the heat involved:

Heat involved = (Enthalpy change) * (moles of O2) Heat involved = (-1652 kJ) * (0.4219 mol)

Heat involved ≈ -696.128 kJ

The negative sign indicates that heat is released in the process. So, approximately 696.128 kJ of heat is involved when 25.6 g of Fe(s) reacts with 13.5 g of O2(g).