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).
Show 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.
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 \( \boxed{420} \, \text{kg} \).
Was this answer helpful?
0
0
Hide Solution
Verified By Collegedunia
Approach Solution -2
Consider the following reaction occurring in the blast furnace:
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 \( \boxed{420} \, \text{kg} \).
