Question

how many liters of bromine are needed to produce 12 moles of Aluminum bromide? The density of bromine is 3.1 m/gl​

Answer 1

Answer:

0.94 L

Explanation:

Step 1: Write the balanced reaction

2 Al + 3 Br₂ = 2 AlBr₃

Step 2: Calculate the moles of Br₂ required to produce 12 moles of AlBr₃

The molar ratio of Br₂ to AlBr₃ is 3:2. Then,

$12molAlBr_3 \times \frac{3molBr_2}{2molAlBr_3} =18molBr_2$

Step 3: Calculate the mass of bromine corresponding to 18 moles

The molar mass of bromine is 159.81 g/mol. Then,

$18mol \times \frac{159.81g}{mol} =2.9 \times 10^{3} g$

Step 4: Calculate the volume of bromine corresponding to 2.9 × 10³ g

The density of bromine is 3.1 g/mL. The volume of bromine is:

$2.9 \times 10^{3} g \times \frac{1mL}{3.1g} \times \frac{1L}{1000mL} = 0.94 L$