Question

Aluminum metal reacts with bromine, a red-brown liquid with a noxious odor. The reaction is vigorous and produces aluminum bromide, a white crystalline substance. A sample of 27g of aluminum yields 266.7g of aluminum bromide. How many grams of bromine react with 18.1g of aluminum?

Answer 1

Answer: The mass of bromine reacted is 160.6 grams.

Step-by-step explanation:

To calculate the number of moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}` .....(1)

Given mass of aluminium = 18.1 g

Molar mass of aluminium = 27 g/mol

Putting values in equation 1, we get:

`\text{Moles of aluminium}=(18.1g)/(27g/mol)=0.670mol`

The chemical equation for the reaction of aluminium and bromide follows:

`2Al+3Br_2\rightarrow 2AlBr_3`

By Stoichiometry of the reaction:

2 moles of aluminium reacts with 3 moles of bromine gas

So, 0.670 moles of aluminium will react with =
`(3)/(2)* 0.670=1.005mol` of bromine gas.

Now, calculating the mass of bromine gas, we use equation 1:

Moles of bromine gas = 1.005 moles

Molar mass of bromine gas = 159.81 g/mol

Putting values in equation 1, we get:

`1.005mol=\frac{\text{Mass of bromine}}{159.81g/mol}\\\\\text{Mass of bromine}=(1.005mol* 159.81g/mol)=160.6g`

Hence, the mass of bromine reacted is 160.6 grams.