Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: artem.metelskiy on March 31, 2011, 06:54:51 PM
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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?
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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.
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When you have a chemical equation, you should try to always balance it and check it if it's balanced.
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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.
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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
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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?
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2(Fe2O3*4H2O) is much better