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The reaction of 60.0 g of aluminum oxide with 30.0 g of carbon produced CO gas and 22.5 g of aluminum metal. What is the percent yield aluminum for this reaction?

Respuesta :

joseaaronlara

Answer:

The answer to your question is: 70.8%

Explanation:

Data

Al₂O₃ = 60 g

C = 30 g

CO = gas

Al = 22.5 g

MW Al₂O₃ = 102 g

MW C = 12 g

MW Al = 54 g

Reaction

                 Al₂O₃    +     3C     ⇒       3 CO    +    2 Al

Limiting reactant

                            102 g of Al₂O₃  -------------- 54 g Al

                             60 g                 --------------   x

                             x = 31.8 g

                            36 g of C ------------------ 54 g of Al

                            30 g of C ------------------  x

                            x = 45 g of Al

Limiting reactant = Al₂O₃

Percent yield = [tex]\frac{22.5}{31.8} x 100[/tex]

Percent yield = 70.75 %

The percent yield of aluminum for this reaction is 70.9%.G

Given:

The reaction of 60.0 g of aluminum oxide with 30.0 g of carbon produced CO gas and 22.5 g of aluminum metal.

To find:

The percent yield of aluminum for this reaction.

Solution:

[tex]Al_2O_3+3C\rightarrow 3CO+2Al[/tex]

Mass of aluminum oxide = 60.0 g

Moles of aluminum oxide =[tex]\frac{60.0g}{101.96 g/mol}=0.588 mol[/tex]

Mass of carbon = 22.5 g

Moles of carbon [tex]= \frac{30.0g}{12.01 g/mol}=2.50 mol[/tex]

According to reaction, 1 mole of aluminum oxide reacts with 3 moles of carbon, then 0.588 moles of aluminum oxide will react with:

[tex]=\frac{3}{1}\times 0.588mol=1.764 \text{mol of C}[/tex]

0.588 moles of aluminum oxide react with 1.764 moles of carbon.

This indicates that:

  • Carbon is an excessive reagent.
  • Aluminum oxide is a limiting reagent.
  • Mass of aluminum produced will depend upon moles of aluminum oxide.

According to reaction, 1 mole of aluminum oxide gives 2 moles of aluminum, then 0.588 moles of aluminum oxide will give:

[tex]=\frac{2}{1}\times 0.588 mol=1.176 \text{ mol of Al}[/tex]

Mass of 1.176 moles of aluminum :

[tex]=1.176 mol\times 26.98 g/mol=31.72 g[/tex]

The theoretical yield of the reaction = 31.73 g

The experimental yield of the reaction = 22.5 g

The percent yield of the reaction :

[tex](Yield)\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100\\\\=\frac{22.5 g}{31.73 g}\times 100=70.91\%\approx70.9\%[/tex]

The percent yield of aluminum for this reaction is 70.9%.

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