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Over the years, the thermite reaction has been used for welding railroad rails, in incendiary bombs, and to ignite solid-fuel rocket motors. The reaction is given below.

Fe2O3(s) + 2 Al(s) -->2 Fe(l) + Al2O3(s)

What masses of iron(III) oxide and aluminum must be used to produce 30.0 g iron? What is the maximum mass of aluminum oxide that could be produced?

OK, I'm stuck on the limiting reagent part...

My work:
30g Fe x (1 mol/55.85g Fe) =0.5372

0.5372 mol Fe x (2 mol Al/ 2 mol Fe) = 0.5372 mol Al

0.5372 mol Al x (26.98 g Al/1 mol)=14.194 g Al

0.5372 mol Fe x (1 mol Fe2O3/2 mol Fe)=.2686mol Fe2O3

.2686 mol Fe2O3 x (55.85*2+16*3) g / 1 mol
=42.895 g Fe2O3

What is the maximum mass of aluminum oxide that could be produced? ---limiting reagent problem?

I take all the masses and convert them to moles and divide by their coefficients
30.0 g Fe = .5372 mol Fe / 2=.2686
14.194 g Al = .5261 mol Al / 2 = .2630
42.895 g Fe2O3 = .2686 / 1 = .2686

Their numbers are pretty similar, probably the only difference is the aluminum, so is that the limiting one?

.5261 mol Al x (1 mol Al2O3/2 mol Al) = .2630 mol Al2O3??

.2630 g Al2O3 x (1 mol/101.96 g)= .002579 g

please help me on this last part.

  • chemistry* -

    This is not a limiting reagent problem.
    The only error I see right off is mass Al needed is closer to 14.5 than to 14.2. I think you just punched the wrong button. As for the maximum amount of Al2O3, just take the moles Fe, convert to moles Al2O3, then to grams and you have it.

  • chemistry* -


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