Chapter 7, Problem 24A

Interpretation Introduction

Interpretation:

The unbalanced chemical equation corresponding to the reaction between barium oxide, calcium oxide and strontium oxide with aluminium needs to be presented.

Concept Introduction:

• A chemical reaction is expressed as a chemical equation having reactants and products on left and right side of the reaction arrow respectively.

  $\text{Reactants }\to \text{ Products}$

• It also gives the physical state (solid (s), liquid (l), gas (g)) and the amount of reactants and products involved in the reaction.

Answer to Problem 24A

  $\text{BaO(s) + Al(s) }\to {\text{Al}}_{\text{2}}{\text{O}}_3\text{(s) + Ba(s)}$

  $\text{CaO(s) + Al(s) }\to {\text{Al}}_{\text{2}}{\text{O}}_3\text{(s) + Ca(s)}$

  $\text{SrO(s) + Al(s) }\to {\text{Al}}_{\text{2}}{\text{O}}_3\text{(s) + Sr(s)}$

Explanation of Solution

i) Barium oxide with aluminium

Reactants:

  $\begin{array}{l}\text{Barium oxide, solid - BaO }(\text{s})\\ \text{Aluminum, solid - Al (s)}\end{array}$

Products:

  $\begin{array}{l}\text{Barium, solid - Ba(s)}\\ {\text{Aluminum oxide, solid - Al}}_{\text{2}}{\text{O}}_{\text{3}}\text{ (s)}\end{array}$

Unbalanced equation:

  $\text{BaO(s) + Al(s) }\to {\text{Al}}_{\text{2}}{\text{O}}_3\text{(s) + Ba(s)}$

ii) Calcium oxide with aluminium

Reactants:

  $\begin{array}{l}\text{Calcium oxide, solid - CaO }(\text{s})\\ \text{Aluminum, solid - Al (s)}\end{array}$

Products:

  $\begin{array}{l}\text{Calcium, solid - Ca(s)}\\ {\text{Aluminum oxide, solid - Al}}_{\text{2}}{\text{O}}_{\text{3}}\text{ (s)}\end{array}$

Unbalanced equation:

  $\text{CaO(s) + Al(s) }\to {\text{Al}}_{\text{2}}{\text{O}}_3\text{(s) + Ca(s)}$

iii) Strontium oxide with aluminium

Reactants:

  $\begin{array}{l}\text{Strontium oxide, solid - SrO }(\text{s})\\ \text{Aluminum, solid - Al (s)}\end{array}$

Products:

  $\begin{array}{l}\text{Strontium, solid - Sr(s)}\\ {\text{Aluminum oxide, solid - Al}}_{\text{2}}{\text{O}}_{\text{3}}\text{ (s)}\end{array}$

Unbalanced equation:

  $\text{SrO(s) + Al(s) }\to {\text{Al}}_{\text{2}}{\text{O}}_3\text{(s) + Sr(s)}$

Conclusion

Therefore, the unbalanced reactions will have the respective group 2 oxides and aluminium metal as reactants with the corresponding metal and aluminium oxide as products.