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Answer :
To determine the additional product that balances the chemical reaction, let's take a look at the reaction:
[tex]\[ H_2SO_4 + 2 \, NaOH \rightarrow Na_2SO_4 + \][/tex]
The products given are:
1. 2 H[tex]\(_2\)[/tex]O
2. 2 OH
3. H[tex]\(_2\)[/tex]O[tex]\(_2\)[/tex]
4. H[tex]\(_3\)[/tex]O
The reaction is between sulfuric acid (H[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex]) and sodium hydroxide (NaOH) to form sodium sulfate (Na[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex]) and water (H[tex]\(_2\)[/tex]O).
### Step-by-step balancing:
1. Balance the Sodium (Na) atoms:
- On the left side, you have 2 Na from 2 NaOH.
- On the right side, to balance the sodium, you have 2 Na in Na[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex].
2. Balance the Sulfate (SO[tex]\(_4\)[/tex]) group:
- On the left side, there is 1 [tex]\(SO_4\)[/tex] in H[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex].
- On the right side, there is 1 [tex]\(SO_4\)[/tex] in Na[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex].
3. Balance the Hydrogen (H) atoms:
- On the left side, there are 2 H from H[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex] and 2 \times 1 H from 2 NaOH, totaling 4 H atoms.
- On the right side, placing 2 H[tex]\(_2\)[/tex]O gives 4 H atoms (2 \times 2 H in H[tex]\(_2\)[/tex]O).
4. Balance the Oxygen (O) atoms:
- On the left side, there are 4 O from H[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex] and 2 \times 1 O from 2 NaOH, totaling 6 O atoms.
- On the right side, there are 4 O in Na[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex] and 2 \times 1 O in 2 H[tex]\(_2\)[/tex]O, also totaling 6 O atoms.
After checking all the elements, you can see that the entire equation is balanced with the addition of 2 H[tex]\(_2\)[/tex]O. Therefore, the additional product that balances the reaction is:
[tex]\[ \boxed{2 \, H_2O} \][/tex]
[tex]\[ H_2SO_4 + 2 \, NaOH \rightarrow Na_2SO_4 + \][/tex]
The products given are:
1. 2 H[tex]\(_2\)[/tex]O
2. 2 OH
3. H[tex]\(_2\)[/tex]O[tex]\(_2\)[/tex]
4. H[tex]\(_3\)[/tex]O
The reaction is between sulfuric acid (H[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex]) and sodium hydroxide (NaOH) to form sodium sulfate (Na[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex]) and water (H[tex]\(_2\)[/tex]O).
### Step-by-step balancing:
1. Balance the Sodium (Na) atoms:
- On the left side, you have 2 Na from 2 NaOH.
- On the right side, to balance the sodium, you have 2 Na in Na[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex].
2. Balance the Sulfate (SO[tex]\(_4\)[/tex]) group:
- On the left side, there is 1 [tex]\(SO_4\)[/tex] in H[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex].
- On the right side, there is 1 [tex]\(SO_4\)[/tex] in Na[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex].
3. Balance the Hydrogen (H) atoms:
- On the left side, there are 2 H from H[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex] and 2 \times 1 H from 2 NaOH, totaling 4 H atoms.
- On the right side, placing 2 H[tex]\(_2\)[/tex]O gives 4 H atoms (2 \times 2 H in H[tex]\(_2\)[/tex]O).
4. Balance the Oxygen (O) atoms:
- On the left side, there are 4 O from H[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex] and 2 \times 1 O from 2 NaOH, totaling 6 O atoms.
- On the right side, there are 4 O in Na[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex] and 2 \times 1 O in 2 H[tex]\(_2\)[/tex]O, also totaling 6 O atoms.
After checking all the elements, you can see that the entire equation is balanced with the addition of 2 H[tex]\(_2\)[/tex]O. Therefore, the additional product that balances the reaction is:
[tex]\[ \boxed{2 \, H_2O} \][/tex]
Thank you for reading the article Which additional product balances the reaction tex H 2SO 4 2NaOH rightarrow Na 2SO 4 tex A tex 2OH tex B tex H 2O 2. We hope the information provided is useful and helps you understand this topic better. Feel free to explore more helpful content on our website!
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