Answer :
To balance the given chemical equation, let's go through the following steps systematically:
1. Write down the unbalanced equation:
[tex]\[ \text{Al} + \text{Conc HNO}_3 (\text{aq}) \rightarrow \text{Al(NO}_3\text{)}_3 + \text{H}_2\text{O} \][/tex]
2. Count the number of each type of atom on both sides of the equation:
- Left side:
- Al: 1
- N: 1
- O: 3 (from HNO[tex]\(_3\)[/tex])
- H: 1
- Right side:
- Al: 1
- N: 3 (from Al(NO[tex]\(_3\)[/tex])_3)
- O: 9 (from Al(NO[tex]\(_3\)[/tex])_3)
- H: 2 (from H[tex]\(_2\)[/tex]O)
3. Balance the atoms one by one:
- To balance nitrogen (N), we notice that there are 3 nitrogen atoms on the right side in Al(NO[tex]\(_3\)[/tex])_3. Therefore, we need 3 molecules of HNO[tex]\(_3\)[/tex] on the left side:
[tex]\[ \text{Al} + 3\text{HNO}_3 \rightarrow \text{Al(NO}_3\text{)}_3 + \text{H}_2\text{O} \][/tex]
- After this adjustment, the counts are as follows:
- Left side:
- Al: 1
- N: 3
- O: 9 (from 3HNO[tex]\(_3\)[/tex])
- H: 3
- Right side:
- Al: 1
- N: 3
- O: 9 (from Al(NO[tex]\(_3\)[/tex])_3)
- H: 2
4. Balance hydrogen atoms (H):
- Now, we have 3 hydrogen atoms on the left and 2 on the right. We need to balance hydrogen by adjusting the number of water molecules. To balance hydrogen, we can introduce 1.5 molecules of water (H[tex]\(_2\)[/tex]O) on the right side:
[tex]\[ \text{Al} + 3\text{HNO}_3 \rightarrow \text{Al(NO}_3\text{)}_3 + \frac{3}{2} \text{H}_2 \text{O} \][/tex]
- This makes the hydrogen count:
- Left side:
- H: 3 (from 3HNO[tex]\(_3\)[/tex])
- Right side:
- H: 3 (1.5 [tex]\(\times\)[/tex] 2 from [tex]\(\frac{3}{2}\)[/tex]H[tex]\(_2\)[/tex]O)
5. Final verification:
The updated equation balances all the atoms. Let's verify:
- Left side:
- Al: 1
- N: 3
- O: 9
- H: 3
- Right side:
- Al: 1
- N: 3
- O: 9 (6 from Al(NO[tex]\(_3\)[/tex])_3, 3 from 1.5 [tex]\(\times\)[/tex] H[tex]\(_2\)[/tex]O)
- H: 3 (1.5 [tex]\(\times\)[/tex] 2 from 1.5 H[tex]\(_2\)[/tex]O)
Thus, we have found that the balanced chemical equation is:
[tex]\[ \text{Al} + 3\text{HNO}_3 \rightarrow \text{Al(NO}_3\text{)}_3 + \frac{3}{2} \text{H}_2\text{O} \][/tex]
This balanced equation shows that for every one aluminum (Al) atom reacting with three molecules of concentrated nitric acid (HNO[tex]\(_3\)[/tex]), one molecule of aluminum nitrate [tex]\(\text{Al(NO}_3\text{)}_3\)[/tex] and one and a half molecules of water (H[tex]\(_2\)[/tex]O) are formed.
1. Write down the unbalanced equation:
[tex]\[ \text{Al} + \text{Conc HNO}_3 (\text{aq}) \rightarrow \text{Al(NO}_3\text{)}_3 + \text{H}_2\text{O} \][/tex]
2. Count the number of each type of atom on both sides of the equation:
- Left side:
- Al: 1
- N: 1
- O: 3 (from HNO[tex]\(_3\)[/tex])
- H: 1
- Right side:
- Al: 1
- N: 3 (from Al(NO[tex]\(_3\)[/tex])_3)
- O: 9 (from Al(NO[tex]\(_3\)[/tex])_3)
- H: 2 (from H[tex]\(_2\)[/tex]O)
3. Balance the atoms one by one:
- To balance nitrogen (N), we notice that there are 3 nitrogen atoms on the right side in Al(NO[tex]\(_3\)[/tex])_3. Therefore, we need 3 molecules of HNO[tex]\(_3\)[/tex] on the left side:
[tex]\[ \text{Al} + 3\text{HNO}_3 \rightarrow \text{Al(NO}_3\text{)}_3 + \text{H}_2\text{O} \][/tex]
- After this adjustment, the counts are as follows:
- Left side:
- Al: 1
- N: 3
- O: 9 (from 3HNO[tex]\(_3\)[/tex])
- H: 3
- Right side:
- Al: 1
- N: 3
- O: 9 (from Al(NO[tex]\(_3\)[/tex])_3)
- H: 2
4. Balance hydrogen atoms (H):
- Now, we have 3 hydrogen atoms on the left and 2 on the right. We need to balance hydrogen by adjusting the number of water molecules. To balance hydrogen, we can introduce 1.5 molecules of water (H[tex]\(_2\)[/tex]O) on the right side:
[tex]\[ \text{Al} + 3\text{HNO}_3 \rightarrow \text{Al(NO}_3\text{)}_3 + \frac{3}{2} \text{H}_2 \text{O} \][/tex]
- This makes the hydrogen count:
- Left side:
- H: 3 (from 3HNO[tex]\(_3\)[/tex])
- Right side:
- H: 3 (1.5 [tex]\(\times\)[/tex] 2 from [tex]\(\frac{3}{2}\)[/tex]H[tex]\(_2\)[/tex]O)
5. Final verification:
The updated equation balances all the atoms. Let's verify:
- Left side:
- Al: 1
- N: 3
- O: 9
- H: 3
- Right side:
- Al: 1
- N: 3
- O: 9 (6 from Al(NO[tex]\(_3\)[/tex])_3, 3 from 1.5 [tex]\(\times\)[/tex] H[tex]\(_2\)[/tex]O)
- H: 3 (1.5 [tex]\(\times\)[/tex] 2 from 1.5 H[tex]\(_2\)[/tex]O)
Thus, we have found that the balanced chemical equation is:
[tex]\[ \text{Al} + 3\text{HNO}_3 \rightarrow \text{Al(NO}_3\text{)}_3 + \frac{3}{2} \text{H}_2\text{O} \][/tex]
This balanced equation shows that for every one aluminum (Al) atom reacting with three molecules of concentrated nitric acid (HNO[tex]\(_3\)[/tex]), one molecule of aluminum nitrate [tex]\(\text{Al(NO}_3\text{)}_3\)[/tex] and one and a half molecules of water (H[tex]\(_2\)[/tex]O) are formed.
