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Consider the reaction 2Al + 6HBr → 2AlBr3 + 3H2. If 8 moles of Al react with 8 moles of HBr, what is the limiting reactant?

A Al
B AlBr3
C H3
D HBr

Respuesta :

Neetoo

Answer:

D) HBr

Explanation:

Given data:

Number of moles of Al = 8 mol

Number of moles of HBr = 8 mol

Limiting reactant = ?

Solution:

2Al + 6HBr      →      2AlBr₃ + 3H₂

now we will compare the moles of reactants with products.

                Al          :         AlBr₃

                   2         :            2

                    8         :          8

                 Al           :          H₂

                 2             :          3

                 8             :           3/2×8 = 12

                 HBr         :         AlBr₃

                   6           :              2

                    8           :            2/6×8 = 2.67

                 HBr         :            H₂

                   6            :            3

                   8             :           3/6×8 = 4

HBr produced less number of moles of product thus it will act as limiting reactant.

The limiting reactant is

D) HBr

Given:

Number of moles of Al = 8 mol

Number of moles of HBr = 8 mol

To find:

Limiting reactant = ?

The balanced chemical equation:

2Al + 6HBr → 2AlBr₃ + 3H₂

On comparing the moles of reactants and products:

Al : AlBr₃

2 : 2

  • Required  moles: 8 : 8

Al : H₂

2  :  3

  • Required moles: 8 : 3/2*8 = 12

HBr : AlBr₃

6 : 2

  • Required moles: 8 : 2/6*8 = 2.67

HBr : H₂

6 : 3

  • Required moles: 8 : 3/6*8 = 4

HBr produces less number of moles of product thus it will act as limiting reactant.

Thus, the correct option is D.

Learn more:

brainly.com/question/13314627

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