Question

In the reaction of aluminum with iron III oxide, if you start with 54.2 grams of aluminum, how many grams of iron III oxide are needed for the reaction to occur? ​

Answer 1

Answer:

$m_{Fe_2O_3}=160.41gFe_2O_3$

Explanation:

Hello!

In this case, since the reaction between aluminum and iron (III) oxide is:

$2Al+Fe_2O_3\rightarrow Al_2O_3+2Fe$

In such way, since there is 2:1 mole ratio between aluminum (atomic mass = 26.98 g/mol) and iron (III) oxide (molar mass = 159.70 g/mol), we'll be able to compute the mass of the required reactant as shown below:

$m_{Fe_2O_3}=54.2gAl*\frac{1molAl}{26.98gAl}* \frac{1molFe_2O_3}{2molAl}*\frac{159.70gFe_2O_3}{1molFe_2O_3}\\\\m_{Fe_2O_3}=160.41gFe_2O_3$

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