Respuesta :
Answer: The maximum amount of magnesium oxide produced during the reaction can be, 3.9 grams.
Explanation : Given,
Mass of [tex]Mg[/tex] = 2.4 g
Mass of [tex]O_2[/tex] = 8.0 g
Molar mass of [tex]Mg[/tex] = 24.3 g/mol
Molar mass of [tex]O_2[/tex] = 31.9 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{2.4g}{24.3g/mol}=0.099mol[/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{8.0g}{31.9g/mol}=0.25mol[/tex]
Now we have to calculate the limiting and excess reagent.
The balanced chemical equation is:
[tex]2Mg+O_2\rightarrow 2MgO[/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, 0.099 moles of [tex]Mg[/tex] react with [tex]\frac{0.099}{2}=0.049[/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.
Now we have to calculate the moles of [tex]MgO[/tex]
From the reaction, we conclude that
As, 2 mole of [tex]Mg[/tex] react to give 2 mole of [tex]MgO[/tex]
So, 0.099 mole of [tex]Mg[/tex] react to give 0.099 mole of [tex]MgO[/tex]
Now we have to calculate the mass of [tex]MgO[/tex]
[tex]\text{ Mass of }MgO=\text{ Moles of }MgO\times \text{ Molar mass of }MgO[/tex]
Molar mass of [tex]MgO[/tex] = 40.3 g/mole
[tex]\text{ Mass of }MgO=(0.099moles)\times (40.3g/mole)=3.9g[/tex]
Therefore, the maximum amount of magnesium oxide produced during the reaction can be, 3.9 grams.
