You must show ALL WORK and INCLUDE UNITS to get full credit.
1. Balance the following equation
___Fe(s) + ___O₂(g) ⟶ ___Fe₂O₃(s)
2. How many grams of iron (III) oxide are formed when 25 grams of iron reacts completely with oxygen?
1. To balance the equation, we need to count the numbers of each type of atom on both sides of the equation and ensure they are equal.
The balanced equation would be:
4Fe(s) + 3O₂(g) ⟶ 2Fe₂O₃(s)
This means that four moles of iron react with three moles of oxygen to produce two moles of iron (III) oxide.
2. We are given 25 grams of iron and need to find out the mass of iron (III) oxide formed. First, we'll convert grams of iron into moles of Iron.
The molar mass of Iron (Fe) is 55.85 g/mol.
Moles of Iron = mass of Iron / molar mass of Iron
Moles of Iron = 25 g / 55.85 g/mol = 0.4477 moles
From the balanced equation, we can see that the molar ratio of Fe to Fe₂O₃ is 4:2 (or 2:1). Therefore, every 2 moles of Iron will produce 1 mole of Fe₂O₃.
Moles of Fe₂O₃ = 0.4477 moles of Fe * (1 mole of Fe₂O₃/ 2 moles of Fe) = 0.2238 moles of Fe₂O₃
Now, we can convert moles of Fe₂O₃ back to grams.
The molar mass of Fe₂O₃ is (2 * 55.85 g/mol for Fe)+(3 * 16.00 g/mol for O) = 159.70 g/mol
Mass of Fe₂O₃ = moles of Fe₂O₃ * molar mass of Fe₂O₃
Mass of Fe₂O₃ = 0.2238 moles * 159.70 g/mol ≈ 35.76 grams
So, when 25 grams of iron reacts completely with oxygen, approximately 35.76 grams of iron (III) oxide are formed.