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Aluminum reacts with bromine, producing aluminum bromide: 2 Al(s) + 3 Br2(l) → 2 AlBr3(s) In a certain experiment, 20.0 mL of bromine (density = 3.10 g/mL) was reacted with excess aluminum to yield 50.3 g of aluminum bromide. What is the percent yield for this experiment?

Respuesta :

Answer : The percent yield for this experiment is, 72.82 %

Explanation :

First we have to calculate the mass of bromine.

[tex]\text{Mass of bromine}=\text{Density of bromine}\times \text{Volume of bromine}[/tex]

[tex]\text{Mass of bromine}=3.10g/ml\times 20.0ml=62g[/tex]

The mass of bromine = 62 g

Now we have to calculate the moles of bromine.

[tex]\text{Moles of }Br_2=\frac{\text{Mass of }Br_2}{\text{Molar mass of }Br_2}=\frac{62g}{159.8g/mole}=0.388mole[/tex]

Now we have to calculate the moles of [tex]C_6H_5COOH[/tex].

The balanced chemical reaction is,

[tex]2Al(s)+3Br_2(l)\rightarrow 2AlBr_3(s)[/tex]

From the balanced reaction, we conclude that

As, 3 moles of [tex]Br_2[/tex] react to give 2 moles of [tex]AlBr_3[/tex]

So, 0.388 moles of [tex]Br_2[/tex] react to give [tex]\frac{2}{3}\times 0.388=0.259[/tex] moles of [tex]AlBr_3[/tex]

Now we have to calculate the mass of [tex]AlBr_3[/tex]

[tex]\text{Mass of }AlBr_3=\text{Moles of }AlBr_3\times \text{Molar mass of }AlBr_3[/tex]

[tex]\text{Mass of }AlBr_3=(0.259mole)\times (266.69g/mole)=69.07g[/tex]

The theoretical yield of [tex]AlBr_3[/tex]  = 69.07 g

The actual yield of [tex]AlBr_3[/tex]  = 50.3 g

Now we have to calculate the percent yield of [tex]AlBr_3[/tex]

[tex]\%\text{ yield of }AlBr_3=\frac{\text{Actual yield of }AlBr_3}{\text{Theoretical yield of }AlBr_3}\times 100=\frac{50.3g}{69.07g}\times 100=72.82\%[/tex]

Therefore, the percent yield for this experiment is, 72.82 %

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