Question

How many grams of Br2 are needed to completely convert 15.0 g Al to AlBr3 ?

Answer 1

Answer:

133 g

Explanation:

Step 1: Write the balanced equation

2 Al(s) + 3 Br₂(l) ⇒ 2 AlBr₃(s)

Step 2: Calculate the moles corresponding to 15.0 g of Al

The molar mass of aluminum is 26.98 g/mol. The moles corresponding to 15.0 g of Al are:

$15.0g \times \frac{1mol}{26.98g} = 0.556mol$

Step 3: Calculate the moles of Br₂ that react with 0.556 moles of Al

The molar ratio of Al to Br₂ is 2:3. The moles of bromine that react with 0.556 moles of aluminum are:

$0.556molAl \times \frac{3molBr_2}{2molAl} = 0.834molBr_2$

Step 4: Calculate the mass corresponding to 0.834 moles of Br₂

The molar mass of bromine is 159.81 g/mol. The mass corresponding to 0.834 moles of Br₂ is:

$0.834mol \times \frac{159.81g}{mol} = 133 g$