6.52 g of oxygen gas reacts with some amount of iron, and 11.50 g of Fe2O3 are formed. What was the amount of iron that reacted with the oxygen?
4Fe + 3O2 ==> 2Fe2O3
mols Fe2O3 = grams/molar mass
Using the coefficients in the balanced equation, convert mols Fe2O3 to mols
Fe in the equation.
You want grams Fe? Convert mols Fe to g. g = mols Fe x atomic mass Fe.
Note that the grams 6.52 O2 is not relevant.
To find the amount of iron that reacted with the oxygen gas, we need to use the concept of stoichiometry. Stoichiometry is the calculation of the quantities of reactants and products in a chemical reaction.
In this reaction, we're given the mass of oxygen gas (6.52 g) and the mass of Fe2O3 formed (11.50 g). We want to find the amount of iron (Fe) that reacted, so we'll assume that all the oxygen reacted with the iron to form Fe2O3.
1. Start by writing the balanced chemical equation for the reaction:
4Fe + 3O2 → 2Fe2O3
2. Determine the molar mass of oxygen (O2) and Fe2O3:
Oxygen (O2): 16.00 g/mol (2 oxygen atoms in O2, each with a molar mass of 16.00 g/mol)
Fe2O3: 159.69 g/mol (2 iron atoms with a molar mass of 55.85 g/mol each and 3 oxygen atoms with a molar mass of 16.00 g/mol each)
3. Convert the given mass of oxygen gas to moles:
moles of O2 = 6.52 g / (16.00 g/mol) = 0.4075 mol
4. Use the balanced equation to relate the moles of O2 to the moles of Fe2O3:
According to the balanced equation, 3 moles of O2 react with 2 moles of Fe2O3.
So, moles of Fe2O3 = (0.4075 mol O2) × (2 mol Fe2O3 / 3 mol O2) = 0.2717 mol Fe2O3
5. Determine the molar mass of Fe2O3 to convert moles of Fe2O3 to grams:
mass of Fe2O3 = (0.2717 mol) × (159.69 g/mol) = 43.45 g
Therefore, the amount of iron that reacted with the oxygen is 43.45 grams.