The thermite reaction, used to weld rails together in the building of railroads, is described by the following equation.
Fe2O3(s) + 2Al(s) --> Al2O3(s) + 2Fe(l)
Calculate the mass of iron metal that can be prepared from 150 grams of aluminum and 250 grams of iron (III) oxide.
250gFe2O3 x (1moleFe2O3 / 85.9995gFe2O3) x (2moleFe / 1moleFe2O3) x (18.99840gFe) = 110gFe
150gAl x (1moleAl / 26.98154gAl) x (2moleFe / 2mole Al) x (18.99840gFe / 1moleFe) = 53gFe
I know the limiting reactant is Al but I do not know how to get the mass of iron metal that is prepared from the aluminum and iron (III) oxide. I know the answer is supposed to be 175gFe. What should I do to get this? Please explain. Thanks
I don't know what you've done. What's the 18.998? Where did you come up with 85.9995? The molar mass of Fe2O3 is 159.7. The atomic mass Fe is 55.85. How do you know Al is the limiting reagent. I think it is Fe2O3.
If you know Al is the limiting reagent, go from there this way.
mols Fe2O3 = 250/159.7 = ?
mols Fe = 2 x that.
grams Fe = mols Fe x 55.85 = 174.85 grams which rounds to 175g.posted by DrBob222
Thank you! I think I understand it a bit betterposted by Anonymous