Question

little confused with this one (c)Calculate the mass of iron metal (in grams) that can be prepared from 150 grams of aluminum and 250 grams of iron(III) oxide

Answer 1

Its a redox reaction: Basically the reaction is as follows:
\[2Al + Fe_{2}O_{3} \rightarrow Al _{2}O _{3}+2Fe\]
There are \[150\div27.0=5.5556\] moles of Aluminium,and
\[250\div(56\times2+16\times3)=1.5625\] moles of iron (III) oxide. Which material is the limiting reagent? Very obviously, iron oxide is the limiting reagent, as there is less of that material.

Thus, the mass of iron that can be prepared will be the amount of iron in 1.5625 mol of iron (III) oxide. I think you will be able to figure that out:)

Answer 2

i have it as 87.25781250009746 but that is wrong :s

Answer 3

Oh sorry I realised where the problem is coming from and apparently we're not supposed to post full solutions so I deleted it:P Sorry!