How many grams of rust are produced from the oxidation of 10.0 lb of iron?
To determine the number of grams of rust produced from the oxidation of 10.0 lb of iron, we need to follow these steps:
Step 1: Convert pounds to kilograms.
To convert pounds to kilograms, we use the conversion factor: 1 lb = 0.453592 kg.
So, 10.0 lb = 10.0 × 0.453592 kg ≈ 4.536 kg.
Step 2: Calculate the molar mass of iron.
The molar mass of iron (Fe) is 55.845 g/mol.
Step 3: Determine the molecular formula of rust.
Rust is formed when iron (Fe) reacts with oxygen (O2) in the presence of water (H2O). The molecular formula of rust is Fe2O3·nH2O, where n represents the number of water molecules.
Step 4: Calculate the molar mass of rust.
The molar mass of Fe2O3 is calculated as follows:
Fe2O3 = 2(Fe) + 3(O) = 2(55.845 g/mol) + 3(16.00 g/mol) ≈ 159.69 g/mol.
Step 5: Calculate the mass of rust using stoichiometry.
The stoichiometric ratio of iron to rust is 2:1 (based on the balanced equation).
Using the equation: (mass of Fe × molar mass of rust) ÷ molar mass of Fe,
we can calculate the mass of rust:
(4.536 kg × 159.69 g/mol) ÷ 55.845 g/mol = 12.996 kg ≈ 12.996 × 1000 g = 12996 g.
Therefore, approximately 12,996 grams (or 13.0 kilograms) of rust are produced from the oxidation of 10.0 lb (or 4.536 kg) of iron.
10 lb x 454 g/lb = ? grams Fe.
mols Fe = grams Fe/atomic mass Fe.
Using the coefficients in the balanced equation, convert mols Fe to mols Fe2O3.
Then convert mols Fe2O3 to g. g = mols x molar mass = ?