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.

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:

$\text{Moles of aluminium}=\frac{18.1g}{27g/mol}=0.670mol$

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 = $\frac{3}{2}\times 0.670=1.005mol$ 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:

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

Hence, the mass of bromine reacted is 160.6 grams.