Question

Consider the following reaction occurring in the blast furnace. \[ {Fe}_3{O}_4(s) + 4{CO}(g) \rightarrow 3{Fe}(l) + 4{CO}_2(g) \] ‘x’ kg of iron is produced when \(2.32 \times 10^3\) kg \(Fe_3O_4\) and \(2.8 \times 10^2 \) kg CO are brought together in the furnace. 
The value of ‘x’ is __________ (nearest integer).

Hint:

In stoichiometry problems involving mass and moles, always make sure to use the correct molar masses and balance the equation to understand the relationships between reactants and products. You can calculate the mass produced using the stoichiometric ratios.

Answer 1

Step 1: Convert Masses to Moles

Moles of Fe\(_3\)O\(_4\) are calculated as: \[ \text{Moles of Fe}_3\text{O}_4 = \frac{2.32 \times 10^6 \, \text{g}}{232 \, \text{g/mol}} = 10^4 \, \text{mol} \] Moles of CO are calculated as: \[ \text{Moles of CO} = \frac{2.8 \times 10^5 \, \text{g}}{28 \, \text{g/mol}} = 10^4 \, \text{mol} \]

Step 2: Identify the Limiting Reactant

From the balanced equation, the stoichiometric ratio of Fe\(_3\)O\(_4\) to CO is 1:4. The available mole ratio is: \[ \frac{10^4 \, \text{mol Fe}_3\text{O}_4}{10^4 \, \text{mol CO}} = 1 \] As the required ratio is 1:4, Fe\(_3\)O\(_4\) is in excess, and CO is the limiting reactant.

Step 3: Calculate the Moles of Fe Produced

According to the balanced equation, 4 moles of CO produce 3 moles of Fe. Therefore, the moles of Fe produced are: \[ \text{Moles of Fe} = \frac{3}{4} \times \text{Moles of CO} = \frac{3}{4} \times 10^4 \, \text{mol} = 7.5 \times 10^3 \, \text{mol} \]

Step 4: Convert Moles of Fe to kg

Using the molar mass of Fe (56 g/mol), the mass of Fe produced is: \[ \text{Mass of Fe} = \text{Moles of Fe} \times \text{Molar Mass of Fe} \] \[ \text{Mass of Fe} = 7.5 \times 10^3 \, \text{mol} \times 56 \, \text{g/mol} = 420 \times 10^3 \, \text{g} = 420 \, \text{kg} \]

Conclusion

The calculated value of \( x \) is \( \boxed{420} \, \text{kg} \).