How many grams of iron can be made from the complete reaction of 21.4 grams of Fe2O3?
Fe2O3 + 3CO -----> 2Fe + 3CO2
mols Fe2O3 = grams/molar mass = 21.5/? = ??? Now convert mols Fe2O3 to mols Fe from the reaction using coefficients in the balanced equation.
??? mols Fe2O3 x (2 mols Fe/1 mol Fe2O3) = xx
Then convert mols Fe x grams Fe.
grams Fe x atomic mass Fe = ?
Post your work if you get stuck.
Oh, I'm glad you asked! Let's see, according to the balanced equation, one mole of Fe2O3 reacts with three moles of CO to form two moles of Fe. The molar mass of Fe2O3 is 159.69 g/mol, so let's calculate the moles of Fe2O3 first.
21.4 g of Fe2O3 / 159.69 g/mol = 0.1342 moles of Fe2O3
Since the ratio of Fe2O3 to Fe is 1:2, this means we'll get double the moles of Fe. So:
0.1342 moles of Fe2O3 x 2 = 0.2684 moles of Fe
Now, let's find the molar mass of Fe:
55.85 g/mol x 0.2684 moles = 14.978 g of Fe
So, approximately 15 grams of iron can be made from the complete reaction of 21.4 grams of Fe2O3. But remember, this is just an approximate answer, so don't take it too seriously!
To determine the number of grams of iron (Fe) that can be made from the complete reaction of 21.4 grams of Fe2O3, we need to perform a stoichiometric calculation.
1. Begin by writing down the balanced chemical equation for the reaction:
Fe2O3 + 3CO → 2Fe + 3CO2
2. Calculate the molar mass of Fe2O3 (in grams per mole):
Fe2 = 2 × atomic mass of Fe = 2 × 55.85 g/mol = 111.7 g/mol
O3 = 3 × atomic mass of O = 3 × 16.00 g/mol = 48.00 g/mol
Total molar mass of Fe2O3 = 111.7 + 48.00 = 159.7 g/mol
3. Use the molar mass of Fe2O3 to convert grams to moles:
moles of Fe2O3 = mass of Fe2O3 / molar mass of Fe2O3
moles of Fe2O3 = 21.4 g / 159.7 g/mol = 0.134 moles
4. Now, use the stoichiometry of the balanced equation to determine the moles of Fe produced. According to the equation, 1 mole of Fe2O3 reacts to form 2 moles of Fe:
moles of Fe = 2 × moles of Fe2O3 = 2 × 0.134 moles = 0.268 moles
5. Finally, calculate the mass of Fe (in grams) produced from the reaction:
mass of Fe = moles of Fe × molar mass of Fe
mass of Fe = 0.268 moles × 55.85 g/mol = 14.95 grams
Therefore, from the complete reaction of 21.4 grams of Fe2O3, approximately 14.95 grams of Fe can be made.
To find out how many grams of iron can be made from the complete reaction of 21.4 grams of Fe2O3, you can use stoichiometry.
Stoichiometry is a method used to calculate the quantities of reactants and products in a chemical reaction based on the balanced equation.
In the given balanced equation:
Fe2O3 + 3CO -----> 2Fe + 3CO2
The coefficient in front of Fe2O3 is 1, which means that 1 mole of Fe2O3 reacts with 3 moles of CO to produce 2 moles of Fe.
To begin, you need to calculate the number of moles of Fe2O3 in 21.4 grams. To do this, use the molar mass of Fe2O3.
The molar mass of Fe2O3 is calculated by adding the atomic masses of the constituent elements:
2(atomic mass of Fe) + 3(atomic mass of O) = 2(55.845 g/mol) + 3(16.00 g/mol) = 159.69 g/mol
Next, divide the given mass of Fe2O3 by its molar mass:
21.4 g / 159.69 g/mol = 0.134 moles of Fe2O3
Since the stoichiometry shows that 1 mole of Fe2O3 reacts with 2 moles of Fe, multiply the number of moles of Fe2O3 by the ratio 2/1:
0.134 moles Fe2O3 x (2 moles Fe / 1 mole Fe2O3) = 0.268 moles of Fe
Finally, convert the number of moles of Fe to grams by multiplying by the molar mass of Fe:
0.268 moles Fe x 55.845 g/mol = 14.99 grams of Fe
Therefore, approximately 14.99 grams of iron can be made from the complete reaction of 21.4 grams of Fe2O3.