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### Topic: Calculating moles and mass problem  (Read 7876 times)

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#### artem.metelskiy

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• Mole Snacks: +0/-0 ##### Calculating moles and mass problem
« on: March 31, 2011, 06:54:51 PM »
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 ##### Re: Calculating moles and mass problem
« Reply #1 on: March 31, 2011, 09:48:29 PM »
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

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« Reply #2 on: April 01, 2011, 02:06:45 AM »
When you have a chemical equation, you should try to always balance it and check it if it's balanced.

#### artem.metelskiy

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« Reply #3 on: April 01, 2011, 02:26:51 AM »
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

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« Reply #4 on: April 01, 2011, 02:49:37 AM »
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
AWK

#### artem.metelskiy

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« Reply #5 on: April 01, 2011, 03:38:32 AM »
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

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« Reply #6 on: April 01, 2011, 04:36:24 AM »
2(Fe2O3*4H2O) is much better
AWK