When oxygen (O) reacts with iron (Fe), it produces rust (Fe2O3). In a closed system, how much oxygen must react with 120 grams of iron to produce 580 grams of rust?

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 molO₂: 15.999 g/molFe₂O₃: 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 massmoles = 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 massmass = 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.

The correct name for n3-ion should be azide.

Hi. where is the graph?