How many grams of CO are needed to react with an excess of Fe2O3 to produce 558 g Fe? The equation for the reaction is: Fe2O3(s) + 2 CO(g) → 3 CO2(g) + 2 Fe(s)
To find the number of grams of CO needed to react with Fe2O3, we can use stoichiometry and the given equation for the reaction.
First, let's find the molar mass of Fe2O3 and CO:
- Molar mass of Fe2O3 = (2 x 55.85 g/mol) + (3 x 16.00 g/mol) = 159.70 g/mol
- Molar mass of CO = 12.01 g/mol + 16.00 g/mol = 28.01 g/mol
Next, let's set up the stoichiometric ratio between Fe2O3 and CO using the coefficients from the balanced equation:
- From the balanced equation, we see that 2 moles of CO react with 1 mole of Fe2O3.
Now, we can set up the conversion factor to find the moles of CO:
- 2 moles CO / 1 mole Fe2O3
To convert from moles to grams, we will use the molar mass of CO:
- 28.01 g CO / 1 mole CO
So, the conversion factor is:
- 2 moles CO / 1 mole Fe2O3 x 28.01 g CO / 1 mole CO = 56.02 g CO / 1 mole Fe2O3
Now, let's calculate the grams of CO:
- 558 g Fe / 1 mole Fe x 1 mole Fe2O3 / 2 moles Fe2O3 x 56.02 g CO / 1 mole Fe2O3 = 15,636.84 g CO
Therefore, approximately 15,636.84 grams of CO are needed to react with an excess of Fe2O3 to produce 558 g Fe.
To determine the grams of CO needed to react with an excess of Fe2O3, we can follow the stoichiometry of the balanced equation.
1. Calculate the molar mass of CO (carbon monoxide):
Carbon (C) = 12.01 g/mol
Oxygen (O) = 16.00 g/mol
Molar mass of CO = (12.01 g/mol) + (16.00 g/mol) = 28.01 g/mol
2. Determine the molar ratio between Fe and CO using the balanced equation:
From the balanced equation, we see that 2 moles of CO are required to produce 2 moles of Fe.
This gives a ratio of 1:1 between CO and Fe.
3. Convert the given mass of Fe to moles:
Grams of Fe = 558 g
Molar mass of Fe = 55.85 g/mol
Moles of Fe = (558 g Fe) / (55.85 g/mol) = 9.99 mol Fe (rounding to 3 significant figures)
4. Use the molar ratio to determine the moles of CO required:
Since the molar ratio is 1:1 between CO and Fe, the moles of CO will also be 9.99 mol.
5. Convert the moles of CO to grams:
Moles of CO = 9.99 mol CO
Molar mass of CO = 28.01 g/mol
Grams of CO = (9.99 mol CO) x (28.01 g/mol) = 279.28 g CO (rounding to 3 significant figures)
Therefore, approximately 279.28 grams of CO are needed to react with an excess of Fe2O3 to produce 558 grams of Fe.