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Aluminum will react with bromine to form aluminum bromide (used as an acid catalyst in organic synthesis). Al(s) + Br2(l) → Al2Br6(s) [unbalanced] How many moles of Al are needed to form 2.43 mol of Al2Br6?

Respuesta :

joseaaronlara

Answer:

The answer to your question is 4.86 moles

Explanation:

Unbalanced Chemical reaction

                       Al(s)  +  Br₂ (l)   ⇒   Al₂Br₆ (s)

               Reactants      Elements      Products

                     1                     Al                    2

                     2                    Br                    6

Balanced chemical reaction

                     2 Al(s)  + 3Br₂ (l)   ⇒   Al₂Br₆ (s)

               Reactants      Elements      Products

                     2                     Al                    2

                     6                    Br                     6

Moles of Al = ?

Moles of Al₂Br₆ = 2.43 moles

- Use proportions to answer this question

              2 moles of Al -------------------- 1 mol of Al₂Br₆

              x                     --------------------  2.43 moles of Al₂Br₆

              x = (2.43 x 2)/1

             x = 4.86 moles of Al

samueladesida43

4.86 moles of Al are needed to form 2.43 mol of Al2Br6.

  • According to this question, aluminum reacts with bromine to form aluminum bromide as follows:

  • Al(s) + Br2(l) → Al2Br6(s)

  • However, this equation is not balanced. The balanced equation using coefficients is as follows:

  • 2Al(s) + 3Br2(l) → Al2Br6(s)

  • According to this equation, 2 moles of Aluminum produces 1 mole of Al2Br6.

  • Hence, 2.43 moles of Al2Br6 will be produced from (2.43 × 2) = 4.86 moles of Al.

  • Therefore, 4.86 moles of Al are needed to form 2.43 mol of Al2Br6.

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