Question

Given below is a chemical reaction. 4Fe + 3O2 -> 2Fe2O3 How much Fe2O3is formed when 5.6 g of Iron reacts completely with oxygen?

• A. 8g
• B. 80g
• C. 160g
• D. 16g

Answer 1

Answer:

A. 8g

Stoichiometry and Molar Mass Calculations

Understanding the Balanced Chemical Equation

• The provided chemical equation is 4Fe + 3O2 → 2Fe2O3.

• This equation indicates that 4 moles of Iron (Fe) react with 3 moles of Oxygen (O₂) to produce 2 moles of Iron(III) Oxide (Fe₂O₃).

• Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in chemical reactions.

Calculating Molar Masses

• The molar mass of Iron (Fe) is approximately 55.845 g/mol. For simpler calculations, it's often rounded to 56 g/mol.

• The molar mass of Iron(III) Oxide (Fe₂O₃) is calculated as: (2 × Molar Mass of Fe) + (3 × Molar Mass of O).

• Using rounded values: (2 × 56 g/mol) + (3 × 16 g/mol) = 112 g/mol + 48 g/mol = 160 g/mol.

Determining Moles of Reactant

• We are given 5.6 g of Iron (Fe).

• To find the number of moles of Fe, we use the formula: Moles = Mass / Molar Mass.

• Moles of Fe = 5.6 g / 56 g/mol = 0.1 moles.

Applying Stoichiometric Ratios

• From the balanced equation, 4 moles of Fe produce 2 moles of Fe₂O₃.

• This means the mole ratio of Fe to Fe₂O₃ is 4:2, or simplified, 2:1.

• Therefore, 0.1 moles of Fe will produce (0.1 moles Fe / 2) = 0.05 moles of Fe₂O₃.

Calculating the Mass of Product

• Now, we convert the moles of Fe₂O₃ back into grams using its molar mass.

• Mass of Fe₂O₃ = Moles of Fe₂O₃ × Molar Mass of Fe₂O₃.

• Mass of Fe₂O₃ = 0.05 moles × 160 g/mol = 8 g.