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You may want to reference (Page) section 4.3 while completing this problem. Magnesium oxide can be made by heating magnesium metal in the presence of the oxygen. The balanced equation for the reaction is 2Mg(s)+O2(g)→2MgO(s) 2 M g ( s ) + O 2 ( g ) → 2 M g O ( s ) When 10.0 g g Mg M g is allowed to react with 10.6 g g O2 O 2 , 12.1 g g MgO Mg O is collected.

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Magnesium oxide can be made by heating magnesium metal in the presence of the oxygen. The balanced equation for the reaction is

[tex]2Mg(s)+O_2(g)\rightarrow 2MgO(s)[/tex]

When 10.0 g Mg is allowed to react with 10.6 g O₂, 12.1 g MgO is collected.

Determine the theoretical yield for the reaction.

Answer : The theoretical yield for the reaction is, 16.66 grams

Explanation : Given,

Mass of Mg = 10 g

Mass of [tex]O_2[/tex] = 10.6 g

Molar mass of Mg = 24 g/mole

Molar mass of [tex]O_2[/tex] = 32 g/mole

Molar mass of MgO = 40 g/mole

First we have to calculate the moles of Mg and [tex]O_2[/tex].

[tex]\text{ Moles of }Mg=\frac{\text{ Mass of }Mg}{\text{ Molar mass of }Mg}=\frac{10g}{24g/mole}=0.4167moles[/tex]

[tex]\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{10.6g}{32g/mole}=0.3312moles[/tex]

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction 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, 0.4167 moles of [tex]Mg[/tex] react with [tex]\frac{0.4167}{2}=0.2084[/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, 0.4167 mole of [tex]Mg[/tex] react to give 0.4165 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]

[tex]\text{ Mass of }MgO=(0.4165moles)\times (40g/mole)=16.66g[/tex]

Thus, the theoretical yield for the reaction is, 16.66 grams

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