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Sagot :
Certainly! Let's walk through the solution step-by-step:
### Step 1: Calculate the number of moles of CO
The mass of CO is given as 2.8 kg, and the molar mass of CO is 28.01 g/mol (which is equivalent to 0.02801 kg/mol).
To find the number of moles of CO:
[tex]\[ \text{Moles of CO} = \frac{\text{Mass of CO}}{\text{Molar mass of CO}} \][/tex]
[tex]\[ \text{Moles of CO} = \frac{2.8 \, \text{kg}}{0.02801 \, \text{kg/mol}} \][/tex]
[tex]\[ \text{Moles of CO} \approx 99.96 \, \text{mol} \][/tex]
### Step 2: Calculate the mass of pure Fe_2O_3
The mass of Fe_2O_3 given is 6.2 kg, but it's only 60% pure.
To find the mass of pure Fe_2O_3:
[tex]\[ \text{Pure Fe}_2\text{O}_3 \text{ mass} = 6.2 \, \text{kg} \times 0.60 \][/tex]
[tex]\[ \text{Pure Fe}_2\text{O}_3 \text{ mass} = 3.72 \, \text{kg} \][/tex]
### Step 3: Calculate the number of moles of Fe_2O_3
The molar mass of Fe_2O_3 is 159.69 g/mol (which is equivalent to 0.15969 kg/mol).
To find the number of moles of Fe_2O_3:
[tex]\[ \text{Moles of Fe}_2\text{O}_3 = \frac{\text{Mass of pure Fe}_2\text{O}_3}{\text{Molar mass of Fe}_2\text{O}_3} \][/tex]
[tex]\[ \text{Moles of Fe}_2\text{O}_3 = \frac{3.72 \, \text{kg}}{0.15969 \, \text{kg/mol}} \][/tex]
[tex]\[ \text{Moles of Fe}_2\text{O}_3 \approx 23.30 \, \text{mol} \][/tex]
### Step 4: Calculate the number of moles of iron produced
The balanced chemical equation for the reaction is:
[tex]\[ 3 \, \text{CO} + \text{Fe}_2\text{O}_3 \longrightarrow 2 \, \text{Fe} + 3 \, \text{CO}_2 \][/tex]
According to the stoichiometric ratio, 1 mole of Fe_2O_3 produces 2 moles of Fe.
To find the number of moles of Fe produced:
[tex]\[ \text{Moles of Fe produced} = \text{Moles of Fe}_2\text{O}_3 \times 2 \][/tex]
[tex]\[ \text{Moles of Fe produced} = 23.30 \, \text{mol} \times 2 \][/tex]
[tex]\[ \text{Moles of Fe produced} \approx 46.59 \, \text{mol} \][/tex]
Therefore, the number of moles of iron produced is approximately 46.59 moles.
### Step 1: Calculate the number of moles of CO
The mass of CO is given as 2.8 kg, and the molar mass of CO is 28.01 g/mol (which is equivalent to 0.02801 kg/mol).
To find the number of moles of CO:
[tex]\[ \text{Moles of CO} = \frac{\text{Mass of CO}}{\text{Molar mass of CO}} \][/tex]
[tex]\[ \text{Moles of CO} = \frac{2.8 \, \text{kg}}{0.02801 \, \text{kg/mol}} \][/tex]
[tex]\[ \text{Moles of CO} \approx 99.96 \, \text{mol} \][/tex]
### Step 2: Calculate the mass of pure Fe_2O_3
The mass of Fe_2O_3 given is 6.2 kg, but it's only 60% pure.
To find the mass of pure Fe_2O_3:
[tex]\[ \text{Pure Fe}_2\text{O}_3 \text{ mass} = 6.2 \, \text{kg} \times 0.60 \][/tex]
[tex]\[ \text{Pure Fe}_2\text{O}_3 \text{ mass} = 3.72 \, \text{kg} \][/tex]
### Step 3: Calculate the number of moles of Fe_2O_3
The molar mass of Fe_2O_3 is 159.69 g/mol (which is equivalent to 0.15969 kg/mol).
To find the number of moles of Fe_2O_3:
[tex]\[ \text{Moles of Fe}_2\text{O}_3 = \frac{\text{Mass of pure Fe}_2\text{O}_3}{\text{Molar mass of Fe}_2\text{O}_3} \][/tex]
[tex]\[ \text{Moles of Fe}_2\text{O}_3 = \frac{3.72 \, \text{kg}}{0.15969 \, \text{kg/mol}} \][/tex]
[tex]\[ \text{Moles of Fe}_2\text{O}_3 \approx 23.30 \, \text{mol} \][/tex]
### Step 4: Calculate the number of moles of iron produced
The balanced chemical equation for the reaction is:
[tex]\[ 3 \, \text{CO} + \text{Fe}_2\text{O}_3 \longrightarrow 2 \, \text{Fe} + 3 \, \text{CO}_2 \][/tex]
According to the stoichiometric ratio, 1 mole of Fe_2O_3 produces 2 moles of Fe.
To find the number of moles of Fe produced:
[tex]\[ \text{Moles of Fe produced} = \text{Moles of Fe}_2\text{O}_3 \times 2 \][/tex]
[tex]\[ \text{Moles of Fe produced} = 23.30 \, \text{mol} \times 2 \][/tex]
[tex]\[ \text{Moles of Fe produced} \approx 46.59 \, \text{mol} \][/tex]
Therefore, the number of moles of iron produced is approximately 46.59 moles.
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