Respuesta :
Answer: The limiting reactants for the reaction is, Mg
Explanation : Given,
Mass of [tex]Mg[/tex] = 10.1 g
Mass of [tex]O_2[/tex] = 10.5 g
Molar mass of [tex]Mg[/tex] = 24 g/mol
Molar mass of [tex]O-2[/tex] = 32 g/mol
First we have to calculate the moles of [tex]Mg[/tex] and [tex]O_2[/tex].
[tex]\text{Moles of }Mg=\frac{\text{Given mass }Mg}{\text{Molar mass }Mg}[/tex]
[tex]\text{Moles of }Mg=\frac{10.1g}{24g/mol}=0.421mol[/tex]
and,
[tex]\text{Moles of }O_2=\frac{\text{Given mass }O_2}{\text{Molar mass }O_2}[/tex]
[tex]\text{Moles of }O_2=\frac{10.5g}{32g/mol}=0.328mol[/tex]
Now we have to calculate the limiting and excess reagent.
The balanced chemical equation is:
[tex]2Mg(s)+O_2(g)\rightarrow 2MgO(s)[/tex]
From the balanced reaction we conclude that
As, 2 mole of [tex]Mg[/tex] react with 1 mole of [tex]O_2[/tex]
So, 0.421 moles of [tex]Mg[/tex] react with [tex]\frac{0.421}{2}=0.2105[/tex] moles of [tex]O_2[/tex]
From this we conclude that, [tex]O_2[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]Mg[/tex] is a limiting reagent and it limits the formation of product.
Hence, the limiting reactants for the reaction is, Mg
