Question

Aluminum reacts with bromine to form aluminum bromide. If 3.00 g of aluminum is mixed with 20.0g of bromine, how many grams of aluminum bromide can be formed?

Answer 1

Molar mass of Al = 26.98 g/mol

mass(Al)= 3.0 g

number of mol of Al,

n = mass of Al/molar mass of Al

=(3.0 g)/(26.98 g/mol)

= 0.1112 mol

Molar mass of Br2 = 159.8 g/mol

mass(Br2)= 20.0 g

number of mol of Br2,

n = mass of Br2/molar mass of Br2

=(20.0 g)/(159.8 g/mol)

= 0.1252 mol

Balanced chemical equation is:

2 Al + 3 Br2 ---> 2 AlBr3

2 mol of Al reacts with 3 mol of Br2

for 0.111193 mol of Al, 0.16679 mol of Br2 is required

But we have 0.125156 mol of Br2

so, Br2 is limiting reagent

we will use Br2 in further calculation

Molar mass of AlBr3,

MM = 1*MM(Al) + 3*MM(Br)

= 1*26.98 + 3*79.9

= 266.68 g/mol

According to balanced equation

mol of AlBr3 formed = (2/3)* moles of Br2

= (2/3)*0.125156

= 0.083438 mol

mass of AlBr3 = number of mol * molar mass

= 8.344*10^-2*2.667*10^2

= 22.25 g

Answer: 22.3 g