Answer:
Limiting reactant is magnesium.
14.1 g remain unreacted.
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:
[tex]n_{Al_2O_3}=120gAl_2O_3*\frac{1molAl_2O_3}{101.96gAl_2O_3}=1.18molAl_2O_3\\\\n_{Mg}=100gMg*\frac{1molMg}{24.31gMg}=4.11molMg[/tex]
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:
[tex]n_{Mg}^{consumed}=1.18molAl_2O_3*\frac{3molMg}{1molAl_2O_3} =3.53molMg[/tex]
Meaning there are more available moles of magnesium than consumed, and therefore it is the limiting reactant. Moreover, it is in excess by:
[tex]n_{Mg}^{excess}=4.11mol-3.53mol=0.58molMg\\\\m_{Mg}^{excess}=0.58molMg*\frac{24.31gMg}{1molMg}\\\\ m_{Mg}^{excess}=14.1gMg[/tex]
It means that 14.1 g of magnesium remain unreacted.
Regards!
