What percentage of iron is contained in an iron ore, which is 60.0% by mass of Fe2O3*note 2 and 3 are subscriptI done one of this problem once but forgot how to do it. Please be detail about the step you take. Don‘t need to provide me answer, I would appreciate it if you can help me set it up.Thanks in advance.
This is basically what my chemistry homework is on tonight. I was just doing a very similar problem! Sorry, but chem is my worst class and I was actually going to ask a question like this haha. So sorry I can't help you. I guess I'll just see what answers you get. But I was thinking these problems are usually solved with that Factor-Label method, right? Ahh never mind i dunno :P AHH here wait, I think I solved it ^_^ I'm sorry if this is wrong. It sounds like kind of a high number. Check this before you turn it in. I found the atomic mass of both Iron (Fe) and Oxygen (O), which are 55.8 g Fe and 16.0 g O. I then multiplied 55.8 g Fe by 2, because there are two Iron atoms in the iron ore compound. Then you multiply 16.0 g O by 3, because of the same reason Fe 55.8 g x 2 111.6 g Fe O 16.0 g x 3 48.0 g Now you have 111.6 g of Iron, and 48.0 g of Oxygen. If you add the two together you get 159.6 g Fe(2)O(3), or your sample of iron ore. Now this is where I get a little doubtful, but I multiplied that number by .6 to get 60% of it. And I got. 95.76 g Fe in iron ore! ) Hope I helped!
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