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).
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).
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Approach Solution - 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, 1 mole of Fe\(_3\)O\(_4\) reacts with 4 moles of CO. The mole ratio of Fe\(_3\)O\(_4\) to CO is 1:4. The available mole ratio is: \[ \frac{10^4}{10^4} = 1 \] Since the reaction requires a ratio of 1:4, Fe\(_3\)O\(_4\) is in excess, and CO is the limiting reactant.
Step 3: Calculate the Moles of Fe Produced
From the balanced equation, 4 moles of CO produce 3 moles of Fe. Therefore: \[ \text{Moles of Fe} = \frac{3}{4} \times \text{Moles of CO} = \frac{3}{4} \times 10^4 = 7.5 \times 10^3 \, \text{mol} \]
Step 4: Convert Moles of Fe to kg
The molar mass of Fe is 56 g/mol, so the mass of Fe 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
Therefore, the value of \( x \) is \( \boxed{420} \, \text{kg} \).
Approach Solution -2
Consider the following reaction occurring in the blast furnace:
\[ \text{Fe}_3\text{O}_4 (s) + 4 \, \text{CO} (g) \rightarrow 3 \, \text{Fe} (l) + 4 \, \text{CO}_2 (g). \]
The given masses are:
- Mass of \( \text{Fe}_3\text{O}_4 = 2.32 \times 10^3 \ \text{kg}\)
- Mass of \(\text{CO} = 2.8 \times 10^2 \ \text{kg}\)
We are asked to find the amount of iron \( x \) (in kg) produced when these amounts of \( \text{Fe}_3\text{O}_4 \) and \( \text{CO} \) are brought together in the furnace.
Step 1: Moles of \( \text{Fe}_3\text{O}_4 \)
The molar mass of \( \text{Fe}_3\text{O}_4 \) is calculated as: \[ \text{Molar mass of Fe}_3\text{O}_4 = 3 \times 56 + 4 \times 16 = 232 \, \text{g/mol}. \] Now, the number of moles of \( \text{Fe}_3\text{O}_4 \) is: \[ \text{Moles of Fe}_3\text{O}_4 = \frac{2.32 \times 10^6 \, \text{g}}{232 \, \text{g/mol}} = 10000 \, \text{mol}. \]
Step 2: Moles of \( \text{CO} \)
The molar mass of \( \text{CO} \) is: \[ \text{Molar mass of CO} = 12 + 16 = 28 \, \text{g/mol}. \] The number of moles of \( \text{CO} \) is: \[ \text{Moles of CO} = \frac{2.8 \times 10^5 \, \text{g}}{28 \, \text{g/mol}} = 10000 \, \text{mol}. \]
Step 3: Identify the Limiting Reagent
The stoichiometric ratio from the balanced equation is: \[ \text{Fe}_3\text{O}_4 : \text{CO} = 1 : 4. \] For 10000 moles of \( \text{Fe}_3\text{O}_4 \), we need: \[ 10000 \times 4 = 40000 \, \text{moles of CO}. \] We have 10000 moles of \( \text{CO} \), which is less than what is required, so \( \text{CO} \) is the limiting reagent.
Step 4: Calculate the Mass of Iron Produced
From the reaction, 4 moles of \( \text{CO} \) produce 3 moles of iron (\( \text{Fe} \)). Therefore, the number of moles of iron produced is: \[ \text{Moles of Fe} = \frac{3}{4} \times 10000 = 7500 \, \text{mol}. \] The molar mass of iron is 56 g/mol. Therefore, the mass of iron produced is: \[ \text{Mass of Fe} = 7500 \times 56 = 420000 \, \text{g} = 420 \, \text{kg}. \]
Final Answer:
The value of \( x \), the mass of iron produced, is \( \boxed{420} \, \text{kg} \).
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