Q84E
Question
Write balanced chemical equations for the following reactions:
(a) metallic aluminum burned in air
(b) elemental aluminum heated in an atmosphere of chlorine
(c) aluminum heated in hydrogen bromide gas
(d) aluminum hydroxide added to a solution of nitric acid
Step-by-Step Solution
Verifieda. Balanced equation of metallic aluminum burned in air is \(4{\rm{Al}}({\rm{s}}) + 3{{\rm{O}}_2}(\;{\rm{g}}) \to 2{\rm{A}}{{\rm{l}}_2}{{\rm{O}}_3}(\;{\rm{s}})\)
b. Balanced equation of elemental aluminum heated in an atmosphere of chlorine is \(2{\rm{Al}}({\rm{s}}) + 3{\rm{C}}{{\rm{l}}_2}(\;{\rm{g}}) \to 2{\rm{AlC}}{{\rm{l}}_3}(\;{\rm{s}})\)
c. Balanced equation of aluminum heated in hydrogen bromide gas is \(2{\rm{Al}}({\rm{s}}) + 3{\rm{B}}{{\rm{r}}_2}(\;{\rm{g}}) \to 2{\rm{AlB}}{{\rm{r}}_3}(\;{\rm{s}})\)
d. Balanced equation of aluminum hydroxide added to a solution of nitric acid is \({\rm{Al}}{({\rm{OH}})_3}(\;{\rm{s}}) + 3{\rm{HN}}{{\rm{O}}_3}({\rm{aq}}) \to {\rm{Al}}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_3}({\rm{aq}}) + 3{{\rm{H}}_2}{\rm{O}}({\rm{l}})\)
- In terms of reaction stoichiometry, a balanced equation depicts the amount of reactants consumed or the amount of products produced.
- Stoichiometry is a mathematical formula that calculates the coefficients of the reactants and products in a chemical process. It denotes the mole ratio at which reactants and products mix to generate products.
- Aluminum Al has a valence electron configuration of \({\rm{3}}{{\rm{s}}^{\rm{2}}}{\rm{3}}{{\rm{p}}^{\rm{1}}}\) and belongs to group IIIA.
Aluminum oxide is formed when aluminium reacts with oxygen. The following is the chemical reaction:
\({\rm{Al}}({\rm{s}}) + {{\rm{O}}_2}(\;{\rm{g}}) \to {\rm{A}}{{\rm{l}}_2}{{\rm{O}}_3}(\;{\rm{s}})\)
The reaction is unbalanced because the number of Al and O atoms in the reactants is 1 and 2, respectively, whereas the number of Al and O atoms in the product is 2 and 3. To make the equation balanced, multiply Al by 4, \({{\rm{O}}_2}\)by 3, and the product \({\rm{A}}{{\rm{l}}_2}{{\rm{O}}_3}\)by 2 to produce the following equation:
\(4{\rm{Al}}({\rm{s}}) + 3{{\rm{O}}_2}(\;{\rm{g}}) \to 2{\rm{A}}{{\rm{l}}_2}{{\rm{O}}_3}(\;{\rm{s}})\)
Aluminum chloride is formed when aluminium reacts with chlorine. The following is the chemical reaction:
\({\rm{Al}}({\rm{s}}) + {\rm{C}}{{\rm{l}}_2}(\;{\rm{g}}) \to {\rm{AlC}}{{\rm{l}}_3}(\;{\rm{s}})\)
The reaction is unbalanced because the reactant has two Cl atoms while the result contains three. Multiply Al by 2, \({\rm{C}}{{\rm{l}}_2}\)by 3, and the resultant \({\rm{AlC}}{{\rm{l}}_3}\)by 2 to get the following equation to balance the equation:
\(2{\rm{Al}}({\rm{s}}) + 3{\rm{C}}{{\rm{l}}_2}(\;{\rm{g}}) \to 2{\rm{AlC}}{{\rm{l}}_3}(\;{\rm{s}})\)
Aluminum bromide is formed when aluminium reacts with bromine. The following is the chemical reaction:
\({\rm{Al}}({\rm{s}}) + {\rm{B}}{{\rm{r}}_2}(\;{\rm{g}}) \to {\rm{AlB}}{{\rm{r}}_3}(\;{\rm{s}})\)
The reaction is unbalanced because there are two Br atoms on the reactant side and three on the product side. To make the equation balanced, multiply Al by 2, 3 and the product AlBr 3 by 2 to get the following equation:
\(2{\rm{Al}}({\rm{s}}) + 3{\rm{B}}{{\rm{r}}_2}(\;{\rm{g}}) \to 2{\rm{AlB}}{{\rm{r}}_3}(\;{\rm{s}})\)
Aluminum nitrate is formed when aluminium hydroxide reacts with nitric acid. The following is the chemical reaction:
\({\rm{Al}}{({\rm{OH}})_3}(\;{\rm{s}}) + {\rm{HN}}{{\rm{O}}_3}({\rm{aq}}) \to {\rm{Al}}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_3}({\rm{aq}}) + {{\rm{H}}_2}{\rm{O}}({\rm{l}})\)
The equation is unbalanced because only the Al atoms on both sides are the same, whereas O, H, and N are not. To balance the reaction, multiply \({\rm{HN}}{{\rm{O}}_3}\)by 3 on the product side and \({{\rm{H}}_2}{\rm{O}}\)by 3 on the reaction side to get the following equation:
\({\rm{Al}}{({\rm{OH}})_3}(\;{\rm{s}}) + 3{\rm{HN}}{{\rm{O}}_3}({\rm{aq}}) \to {\rm{Al}}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_3}({\rm{aq}}) + 3{{\rm{H}}_2}{\rm{O}}({\rm{l}})\)