Question

The thermite reaction occurs when iron(III) oxide reacts with solid

aluminum. The reaction is so hot that molten iron forms as a product.
>
Fe2O3(s) + Al(s) → Fe(C) + Al2O3(s)
What mass of aluminum should be used in order to completely
consume 10.0 g Fe2O3(s)? If the reaction described produces 5.3 g
Al2O3(s), what is the percent yield?

Answer 1

`m_(Al)=3.38gAl`

`Y=83.1\%`

Step-by-step explanation:

Hello.

In this case, for the given balanced reaction:

`Fe_2O_3(s) + 2Al(s)\rightarrow 2Fe(s) + Al_2O_3(s)`

For 10.0 g of iron (III) oxide (molar mass = 160 g/mol), based on the 1:2 mole ratio with Al (atomic mass = 27 g/mol), the required mass is then:

`m_(Al)=10.0gFe_2O_3*(1molFe_2O_3)/(160gFe_2O_3) *(2molAl)/(1molFe_2O_3) *(27gAl)/(1molAl) \\\\m_(Al)=3.38gAl`

Moreover, as 5.3 g of aluminum oxide are actually yielded, from the 10.0 g of iron (III) oxide, we can compute the theoretical mass of aluminum oxide (molar mass = 102 g/mol) via their 1:1 mole ratio:

`m_(Al_2O_3)=10.0gFe_2O_3*(1molFe_2O_3)/(160gFe_2O_3) *(1molAl_2O_3)/(1molFe_2O_3) *(102gAl_2O_3)/(1molAl_2O_3) \\\\m_(Al_2O_3)=6.38gAl_2O_3`

Thus, the percent yield (actual/theoretical*100%) turns out:

`Y=(5.3 g)/(6.38 g)* 100\%\\\\Y=83.1\%`

Best regards.