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.
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
ALUMINUM:
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
IRON(III) OXIDE
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.
2 answers
thanks!