I need help with a chemistry problem...Please help!?

Question

I've tried and tried, but I just can't seem to get this...Please help!

The problem:
4Fe+3O3--->2Fe2O3

An iron nail with a mass of 2.7 g was left to rust. 47% of the iron rusted.
a.) Calculate the mass of iron (III) oxide formed.

b.) Calculate the final mass of the rusty nail.

Answer 1

First of all rust is Iron Oxide Fe2O3.

The given reactant Fe is 2.7g but not all was used. Only 47% of 2.7g was used in the reaction.

Therefore, Fe used is .47 x 2.7 = 1.269g

convert 1.268 g Fe into moles

1.268 g Fe x [ 1 mole / 56 g ] unit g cancels out and you have 0.0226 mole Fe reacted.

a) Convert moles Fe into moles of Fe2O3
0.0226 m Fe x [ 2 moles Fe2O3 / 4 m Fe ] = 0.011 moles of Fe2O3 produced

Convert 0.011 moles Fe2O3 into g of Fe2O3.... this is your answer... you will do the conversion from moles into g yourself using the same setup we used above. Except that ypou like to cancel out moles this time, so put moles on the bottom and g of Fe2O3 on top/denominator.

b) This is little tricky-you need to use logic.

Rusty Nail = Fe2O3 + Fe

So, whatever g of Fe2O3 we got from a) is Fe2O3 and add it to the remainder of Fe that was NOT used in the reaction.
remember Fe used was 47%, so NOT used was 53% which equals 2.7g x .53 = 1.431 g

So add g from a) + 1.431g = xx g is your answer

Answer 2

wow. well, start by compiling all the little things you know.
personally, i remembered this equation when i read the problem:
d=m/v; density=mass/volume
honestly, though, i'm not sure where to go with that... good luck! try to ask your teacher, or a friend!

Answer 3

Why do you want to calculate just scrape off the rust and weigh it SIMPLE

Answer 4

Its easy. you need the atomic weights of Fe and O. You better to contact me via google talk, MSN messenger or Skype so I can teach you how to do it.

gtalk - [e-mail]
Skype - chathuranga.chandrasekara
MSN - [e-mail]

Thanks