Question

Aluminum bromide can be prepared by the reaction of aluminum metal with bromine gas shown by the equation:2 Al + 3 Br2 --> 2 AlBr3Now suppose that 5.6 g of aluminum reacts with 4.4 g of bromine. Calculate the following:A. Calculate the mass of aluminum bromide that can be produced from 5.6 g of Al.B. Calculate the mass of aluminum bromide that can be produced from 4.4 g of bromine.C. Calculate the percent yield if the actual yield was 3.4 g AlBr3.

Answer 1

We are given the balanced formula for the reaction so we can continue with the calculations.

A. To calculate the mass of AlBr3 formed from 5.6 g of Al we will do the following calculations:

1. We find the moles corresponding to 5.6 grams of Al. To do this, we divide the 5.6 grams by the molar mass of aluminum. The molar mass of Al is 26.98g/mol.

2. We find the moles of AlBr3 produced. From stoichiometry we know that two moles of Al produce two moles of AlBr3, so the AlBr3/Al ratio is 2/2.

3. We find the grams of AlBr3 by multiplying the moles of AlBr3 found by the molar mass. The molar mass of AlBr3 is:266.69g/mol.

So, the grams of AlBr3 formed from 5.6 g of Al will be:

`5.6gAl*(1molAl)/(MolarMass,gAl)*(2molAlBr_3)/(2molAl)*(MolarMass,gAlBr_3)/(1molAlBr_3)`
`5.6gAl*(1molAl)/(26.98gAl)*(2molAlBr_3)/(2molAl)*(266.69gAlBr_3)/(1molAlBr_3)=55.3gAlBr_3`

Answer A. the mass of aluminum bromide that can be produced from 5.6 g of Al is 55.3 grams.

B. To calculate the mass of AlBr3 produced from 4.4 grams of Br2 we follow steps 1, 2, and 3 described above.

In step 1, we will use the molar mass of Br. The molar mass of Br2 is 159.808g/mol.

In step 2, we will taka in count that the ratio AlBr3 to Br2 is 2/3

So, the mass of AlBr3 will be:

`4.4gBr*(1molBr_2)/(MolarMass,gBr_2)*(2molAlBr_3)/(3molBr_2)*(MolarMass, gAlBr_(3))/(1molAlBr_(3))`
`4.4gBr*(1molBr_2)/(159.808gBr_2)*(2molAlBr_3)/(3molBr_2)*(266.69gAlBr_3)/(1molAlBr_3)=4.89gAlBr_3`

Answer B. the mass of aluminum bromide that can be produced from 4.4 g of Br is 4.9 grams.

C. The percent yield will be found with the following reaction:

`PercentYield=(ActualYield)/(TheoreticalYield)*100\%`

We will use the minor value of AlBr3 found. It means we will use as theoretical yield 4.9 grams of AlBr3.

`\begin{gathered} PercentYield=(3.4gAlBr_3)/(4.9gAlBr_3)*100\% \\ PercentYield=69\% \end{gathered}`

Answer C. The percent yield is 69%

In summary, we will have:

A. The mass of aluminum bromide that can be produced from 5.6 g of Al is 55.3 grams.

B. The mass of aluminum bromide that can be produced from 4.4 g of Br is 4.9 grams.

C. The percent yield is 69%