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1. Suppose 0.7542 g of magnesium reacts with excess oxygen to form magnesium oxide as the only product, what would be the theoretical yield of the product?


2. If 0.8922 g of magnesium oxide is obtained from the reaction indicated in #1 above, what would be the percent yield of the magnesium oxide?




3. Suppose the percent yield was calculated to be over 100%, what possible reasons could you give to account for it?

Respuesta :

Answer :

(1) The theoretical yield of product, [tex]MgO[/tex] is, 1.257 grams.

(2) The percent yield of [tex]MgO[/tex] is, 64.13 %

(3) If the percent yield is calculated to be over 100% then there might be some impurity present in the desired product.

Solution : Given,

Mass of Mg = 0.7542 g

Molar mass of Mg = 24 g/mole

Molar mass of MgO = 40 g/mole

First we have to calculate the moles of Mg.

[tex]\text{ Moles of }Mg=\frac{\text{ Mass of }Mg}{\text{ Molar mass of }Mg}=\frac{0.7542g}{24g/mole}=0.03142moles[/tex]

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

The balanced chemical reaction is,

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

From the reaction, we conclude that

As, 2 mole of [tex]Mg[/tex] react to give 2 mole of [tex]MgO[/tex]

So, 0.03142 moles of [tex]Mg[/tex] react to give 0.03142 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=(0.03142moles)\times (40g/mole)=1.257g[/tex]

Theoretical yield of [tex]MgO[/tex] = 1.257 g

Experimental yield of [tex]MgO[/tex] = 0.8922 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{0.8922g}{1.257g}\times 100=70.97\%[/tex]

Therefore, the percent yield of [tex]MgO[/tex] is, 70.97 %

If the percent yield is calculated to be over 100% then the product would be greater than 1.257 g which indicates that there might be some impurity present in the desired product.

pstnonsonjoku

The percent yield is 71.95%. A percent yield above 100% implies the presence of an impurity such as Mg3N2.

The reaction occurs as follows;

2Mg + O2 ----> 2MgO

Number of moles of Mg = 0.7542 g/24 g/mol = 0.031 moles

Since oxygen is in excess, Mg is the limiting reactant.

From the reaction equation;

2 moles of Mg yields 2  mole of MgO

0.031 moles of Mg yields 0.031 moles × 2 mole/2 moles  = 0.031 moles

Theoretical yield of MgO =  0.031 moles × 40 g/mol = 1.24 g

Percent yield = actual yield/theoretical yield  × 100/1

Actual yield = 0.8922 g

Percent yield =  0.8922 g /1.24 g  × 100/1

= 71.95%

If the percent yield was calculated to be over 100% then an impurity such as Mg3N2 may be present.

Learn more: https://brainly.com/question/4147359

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