Question

H2SO4 + NaOH → Na2SO4 + H2O

Balance this reaction if necessary.
Responses
A balancedbalanced
B 1, 2, 1, 21, 2, 1, 2
C 1, 2, 1, 41, 2, 1, 4
D 1, 2, 2, 2

Answer 1

To balance the reaction \( \text{H}_2\text{SO}_4 + \text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + \text{H}_2\text{O} \), we first count the number of atoms of each element on both sides of the equation.

The unbalanced equation has:
- On the left:
- H: 2 (from \(\text{H}_2\text{SO}_4\)),
- S: 1 (from \(\text{H}_2\text{SO}_4\)),
- O: 4 (from \(\text{H}_2\text{SO}_4\)) + 1 (from \(\text{NaOH}\) for each NaOH that reacts) = 4 + x,
- Na: x.

- On the right:
- Na: 2 (from \(\text{Na}_2\text{SO}_4\)),
- S: 1 (from \(\text{Na}_2\text{SO}_4\)),
- O: 4 (from \(\text{Na}_2\text{SO}_4\)) + 1 (from \(\text{H}_2\text{O}\)) = 4 + 1,
- H: 2 (from \(\text{H}_2\text{O}\)).

To balance:
1. There are 2 Na on the right side, so we need 2 Na on the left side, which means we need 2 \(\text{NaOH}\):
\[
\text{H}_2\text{SO}_4 + 2 \text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2 \text{H}_2\text{O}
\]

Now, let’s count the atoms again:
- Left:
- H: 2 (from \(\text{H}_2\text{SO}_4\)) + 2 (from \(2 \text{NaOH}\)) = 4,
- S: 1,
- O: 4 (from \(\text{H}_2\text{SO}_4\)) + 2 (from \(2 \text{NaOH}\)) = 6,
- Na: 2.

- Right:
- Na: 2,
- S: 1,
- O: 4 (from \(\text{Na}_2\text{SO}_4\)) + 2 (from \(2 \text{H}_2\text{O}\)) = 6,
- H: 4 (from \(2 \text{H}_2\text{O}\)).

Now both sides have the same number of each atom:
- H: 4
- Na: 2
- S: 1
- O: 6

The balanced equation is:
\[
\text{H}_2\text{SO}_4 + 2 \text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2 \text{H}_2\text{O}
\]

So, the correct choice representing the coefficients is:
**D 1, 2, 1, 2**.