Question

If 120 grams of aluminum oxide reacts with 100 grams of magnesium, which reactant is in excess and by how much?

Al2O3 + 3Mg —> 3MgO + 2Al

Answer 1

Limiting reactant is magnesium.

14.1 g remain unreacted.

Step-by-step explanation:

Hello!

In this case, given the balanced chemical reaction, we are able to compute the moles of each reactant by using their molar masses:

`n_(Al_2O_3)=120gAl_2O_3*(1molAl_2O_3)/(101.96gAl_2O_3)=1.18molAl_2O_3\\\\n_(Mg)=100gMg*(1molMg)/(24.31gMg)=4.11molMg`

Now, since they are reacting based on a 1:3 mole ratio, we can compute the moles of magnesium consumed by the 1.18 moles of aluminum oxide as follows:

`n_(Mg)^(consumed)=1.18molAl_2O_3*(3molMg)/(1molAl_2O_3) =3.53molMg`

Meaning there are more available moles of magnesium than consumed, and therefore it is the limiting reactant. Moreover, it is in excess by:

`n_(Mg)^(excess)=4.11mol-3.53mol=0.58molMg\\\\m_(Mg)^(excess)=0.58molMg*(24.31gMg)/(1molMg)\\\\ m_(Mg)^(excess)=14.1gMg`

It means that 14.1 g of magnesium remain unreacted.

Regards!