Question

If 1.485 grams of Al is mixed with 8.797 grams of Br2 and allowed to react according to the balanced equation: 2 Al(s) + 3 Br2(1)Al2Br6(s) A. What is the limiting reagent? OBr2 ΟΑΙ B. How many grams of Al2Bre could be produced? grams Al2Br6 . C. What mass, in grams, of the excess reagent will remain? grams of excess reagent

Answer 1

A)

Molar mass of Al = 26.98 g/mol

mass(Al)= 1.485 g

use:

number of mol of Al,

n = mass of Al/molar mass of Al

=(1.485 g)/(26.98 g/mol)

= 5.504*10^-2 mol

Molar mass of Br2 = 159.8 g/mol

mass(Br2)= 8.797 g

use:

number of mol of Br2,

n = mass of Br2/molar mass of Br2

=(8.797 g)/(1.598*10^2 g/mol)

= 5.505*10^-2 mol

Balanced chemical equation is:

2 Al + 3 Br2 ---> Al2Br6

2 mol of Al reacts with 3 mol of Br2

for 5.504*10^-2 mol of Al, 8.256*10^-2 mol of Br2 is required

But we have 5.505*10^-2 mol of Br2

so, Br2 is limiting reagent

Answer: Br2

B)

Molar mass of Al2Br6,

MM = 2*MM(Al) + 6*MM(Br)

= 2*26.98 + 6*79.9

= 533.36 g/mol

According to balanced equation

mol of Al2Br6 formed = (1/3)* moles of Br2

= (1/3)*5.505*10^-2

= 1.835*10^-2 mol

use:

mass of Al2Br6 = number of mol * molar mass

= 1.835*10^-2*5.334*10^2

= 9.787 g

Answer: 9.79 g

C)

According to balanced equation

mol of Al reacted = (2/3)* moles of Br2

= (2/3)*5.505*10^-2

= 3.67*10^-2 mol

mol of Al remaining = mol initially present - mol reacted

mol of Al remaining = 5.504*10^-2 - 3.67*10^-2

mol of Al remaining = 1.834*10^-2 mol

Molar mass of Al = 26.98 g/mol

use:

mass of Al,

m = number of mol * molar mass

= 1.834*10^-2 mol * 26.98 g/mol

= 0.4948 g

Answer: 0.495 g