Answer:
- It is not possible to produce 580 grams of rust with 120 grams of iron.
Explanation:
Here, I show you why it is not possible to produce 580 grams of rust with 120 grams of iron.
A. Molar masses of the substances:
- Fe: 55.845g/mol 2.1488 mol
- Fe₂O₃: 159,69 g/mol 3.6320 mol
B. The chemical equation is:
4Fe + 3O₂ → 2Fe₂O₃
How many grams of iron are there in 580 grams of rust (Fe₂O₃) ?
1. Convert the mass of Fe₂O₃ into number of moles:
- moles = mass in grams / molar mass
- moles = 580g / (159.69 g/mol) = 3.6320 mol of Fe₂O₃
2. Calculate the number of moles of Fe in 3.6320 mol of Fe₂O₃
There are 2 moles of Fe per mole of Fe₂O₃:
- 2 mol Fe / mol Fe₂O₃ × 3.6320 mol Fe₂O₃ = 7.2640 mol Fe
3. Calculate the mass of 7.2640 moles of Fe:
- mass = number of moles × atomic mass
- mass = 7.2640 mol × 55.845 g/mol = 405.66 g
Thus, 120 grams of Fe is too little to produce the mentioned amount of rust; you need about 406 grams of Fe to produce 580 grams of rust.
