To identify compounds [tex]\(X\)[/tex] and [tex]\(Y\)[/tex] in the given reaction, follow these steps:
1. Write the Unbalanced Reaction:
[tex]\[
2 \textrm{Al(OH)}_3 + 3 \textrm{H}_2\textrm{SO}_4 \rightarrow X + 6Y
\][/tex]
2. Identify Possible Products:
- Aluminum hydroxide [tex]\( \textrm{Al(OH)}_3 \)[/tex] reacts with sulfuric acid [tex]\( \textrm{H}_2\textrm{SO}_4 \)[/tex].
- Typically, this reaction forms aluminum sulfate and water.
3. Determine the Formula for [tex]\(X\)[/tex]:
- The sulfate ion is [tex]\( \textrm{SO}_4^{2-} \)[/tex].
- Aluminum forms a [tex]\( \textrm{Al}^{3+} \)[/tex] ion.
- Combining [tex]\( \textrm{Al}^{3+} \)[/tex] with [tex]\( \textrm{SO}_4^{2-} \)[/tex] forms [tex]\( \textrm{Al}_2(\textrm{SO}_4)_3 \)[/tex] to balance the charges.
4. Confirm the Formula for [tex]\(Y\)[/tex]:
- Water is commonly formed when hydroxides react with acids.
- [tex]\(Y\)[/tex] is [tex]\( \textrm{H}_2\textrm{O}\)[/tex].
5. Write the Balanced Equation:
[tex]\[
2 \textrm{Al(OH)}_3 + 3 \textrm{H}_2\textrm{SO}_4 \rightarrow \textrm{Al}_2(\textrm{SO}_4)_3 + 6 \textrm{H}_2\textrm{O}
\][/tex]
6. Verify the Balancing:
- There are 2 aluminum atoms on both sides.
- There are 6 oxygen atoms from the hydroxides and 12 from the sulfates, totaling 18 oxygens on the left side. On the right, there are 12 from [tex]\( \textrm{Al}_2(\textrm{SO}_4)_3 \)[/tex] and 6 from [tex]\( 6 \textrm{H}_2\textrm{O} \)[/tex], totaling 18 oxygens, so it balances.
- Hydrogen atoms total 12 on both sides (6 from [tex]\(2 \textrm{Al(OH)}_3\)[/tex] and 6 from [tex]\(\textrm{H}_2\textrm{SO}_4\)[/tex]).
- Sulfate groups balance with 3 sulfates on each side.
The correct products are:
- [tex]\(X = \textrm{Al}_2(\textrm{SO}_4)_3\)[/tex]
- [tex]\(Y = \textrm{H}_2\textrm{O}\)[/tex]
So, the answer is:
[tex]\[ X = \textrm{Al}_2(\textrm{SO}_4)_3; \, Y = \textrm{H}_2\textrm{O} \][/tex]
