[artem.metelskiy]

I need some help to solve this problem:
Iron reacts slowly with oxygen and water to form a compound commonly called rust (Fe₂O₃ * 4H₂O) For 45.2 kg of rust. Calculate: a) the moles of compound; b) The moles of Fe₂O₃; c) the grams of Iron.

I'm not sure where to start, should I frist balance the chemical equation Fe+O₂+H₂O -> Fe₂O₃ * 4H₂O?

[DevaDevil]

Balancing is always a good thing to do.

for the number of moles you first will need to calculate the molar mass of rust.

for the number of grams of iron you will need your balanced equation.

[opti384]

When you have a chemical equation, you should try to always balance it and check it if it's balanced.

[artem.metelskiy]

for part a) Molar mass of rust is 159.7 so 4520g/159.7=28.30


for the rest of problem I don't know how to balance this equation when it has Fe₂O₃ * 4H₂O part in it. Can someone please explain? Thanks.

[AWK]

45.2 kg = 45200 g
Molar mass of rust is 157.9 + 4x18
Your calculations are based on formula (of hydrate) and there is no need for using balancing of reaction (though it may be helpful in this way)
Fe₂O₃^.4H₂O= Fe₂O₃ or (2Fe + 3/2O₂) + 4H₂O

[artem.metelskiy]

Fe₂O₃^.4H₂O= Fe₂O₃ or (2Fe + 3/2O₂) + 4H₂O

Since we don't want to have 1/2 coefficient in front of O₂, should I multiply the whole equation by 2?
4Fe+3O₂+8H₂O->2Fe₂O₃*8H₂O
Would that be a correct balanced equation?

[AWK]

2(Fe₂O₃*4H₂O) is much better