Magnesium oxide can be made by heating magnesium metal in the presence of the oxygen. The balanced equation for the reaction is 2 Mg(s) + O2(g) → 2 MgO(s) Now consider that you react 10.0 g Mg with 6.00 g O2 gas. If you were able to collect 9.62 g of MgO, what would be your percent yield for the reaction?

From this we conclude that, [tex]Mg[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]O_2[/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, 1 mole of [tex]O_2[/tex] react to give 2 mole of [tex]MgO[/tex]

So, 0.1875 moles of [tex]O_2[/tex] react to give [tex]0.1875\times 2=0.375[/tex] moles 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=(0.375moles)\times (40g/mole)=15g[/tex]

Theoretical yield of [tex]MgO[/tex] = 15 g

Experimental yield of [tex]MgO[/tex] = 9.62 g

Now we have to calculate the percent yield of [tex]MgO[/tex]

[tex]\% \text{ yield of }MgO=\frac{\text{ Experimental yield of }MgO}{\text{ Theretical yield of }MgO}\times 100[/tex]

[tex]\% \text{ yield of }MgO=\frac{9.62g}{15g}\times 100=64.13\%[/tex]

Therefore, the percent yield of [tex]MgO[/tex] is, 64.13 %

shubhamchouhanvrVT shubhamchouhanvrVT · Answer 2 · 2022-03-28T11:23:38+02:00

The percentage yield of the Magnesium oxide MgO will be 64.13 %

What will be the percentage yield of the Magnesium Oxide?

It is given that in the question that:-

Mass of Mg = 10 g

Mass of [tex]O_2[/tex] = 6 g

Molar mass of Mg = 24 g/mole

The 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]\rm Moles \ of\ Mg= \dfrac{Maas \ of\ Mg}{Molar\ mass\ of\ Mg}= \dfrac{10}{24} =0.4167 moles[/tex]

[tex]\rm Moles \ of\ O_2= \dfrac{Maas \ of\ O_2}{Molar\ mass\ of\ O_2}= \dfrac{6}{32} =0.1875\ moles[/tex]

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

[tex]\rm 2Mg+O_2--- > 2MgO[/tex]

From the balanced reaction, we conclude that

As 1 mole of [tex]O_2[/tex] reacting with [tex]0.1875\times2=0.375[/tex] 2 mole Mg of

So, 0.1875 moles of reacting with [tex]0.1875\times2=0.375[/tex] moles of Mg

From this, we conclude that Mg is an excess reagent because the given moles are greater than the required moles and [tex]O_2[/tex] is a limiting reagent and it limits the formation of the product.

Now we have to calculate the moles of MgO

From the reaction, we conclude that

As 1 mole of reacting to give 2 moles of

So, 0.1875 moles of [tex]O_2[/tex] reacting to give moles of MgO

Now we have to calculate the mass of

[tex]\rm Mass \ of \ MgO = Moles\ of\ MgO\times Molar\ mass\ of\ MgO[/tex]

[tex]\rm Mass \ of\ MgO=0.375\times 40=15\ g[/tex]

Theoretical yield of = 15 g

The experimental yield of = 9.62 g

Now we have to calculate the percent yield of

% yield of MgO

[tex]=\dfrac{Experimental \ yield}{Theoretical \ Yield} \times 100[/tex]

[tex]=\dfrac{0.62}{15} \times 100=64.13[/tex]

Thus the percentage yield of the Magnesium oxide MgO will be 64.13 %

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