Iron (III) oxide is formed when iron combines with oxygen in the air. How many grams of Fe2O3 are formed when 16.7 g of Fe reacts completely with oxygen?
a. 12.0g
b. 23.9g
c. 47.8g
d. 95.6g
Write the equation and balance it.
Convert 16.7 g Fe to mols. mol = grams/molar mass
Use the coerfficients in the balanced equation to convert mols Fe to mols Fe2O3.
Now convert mols Fe2O3 to grams. g = mols x molar mass.
To calculate the grams of Fe2O3 formed when 16.7 g of Fe reacts completely with oxygen, you need to use Stoichiometry.
Stoichiometry is a mathematical method used to calculate the quantities of reactants and products involved in a chemical reaction. It involves using balanced chemical equations and converting between moles and grams.
First, let's write the balanced chemical equation for the reaction:
4 Fe + 3 O2 -> 2 Fe2O3
From the balanced equation, we can see that 4 moles of Fe react with 3 moles of O2 to produce 2 moles of Fe2O3.
To solve the problem, follow these steps:
Step 1: Convert the given mass of Fe to moles.
Molar mass of Fe = 55.845 g/mol
Moles of Fe = Mass of Fe / Molar mass of Fe
Moles of Fe = 16.7 g / 55.845 g/mol
Step 2: Use the stoichiometric ratio to find moles of Fe2O3.
Since the stoichiometry is 4:2 (4 moles of Fe react to produce 2 moles of Fe2O3), we can say:
2 moles of Fe2O3 = 4 moles of Fe
Moles of Fe2O3 = (Moles of Fe) x (2 moles of Fe2O3 / 4 moles of Fe)
Step 3: Convert moles of Fe2O3 to grams.
Molar mass of Fe2O3 = 159.6882 g/mol
Mass of Fe2O3 = Moles of Fe2O3 x Molar mass of Fe2O3
Calculating:
Moles of Fe = 16.7 g / 55.845 g/mol = 0.2984 mol
Moles of Fe2O3 = (0.2984 mol) x (2 mol Fe2O3 / 4 mol Fe) = 0.1492 mol
Mass of Fe2O3 = 0.1492 mol x 159.6882 g/mol = 23.8 g
Therefore, the answer is b. 23.9g.