To balance the chemical equation \(\text{H}_2\text{SO}_4 + \text{NaNO}_2 \rightarrow \text{HNO}_2 + \text{Na}_2\text{SO}_4\), we need to ensure that the number of atoms of each element is the same on both sides.
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Write down the unbalanced equation:
\[ \text{H}_2\text{SO}_4 + \text{NaNO}_2 \rightarrow \text{HNO}_2 + \text{Na}_2\text{SO}_4 \]
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List the elements and their counts on both sides:
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Left side:
- H: 2 (from \(\text{H}_2\text{SO}_4\))
- S: 1 (from \(\text{H}_2\text{SO}_4\))
- O: 4 (from \(\text{H}_2\text{SO}_4\)) + 2 (from \(\text{NaNO}_2\)) = 6
- Na: 1 (from \(\text{NaNO}_2\))
- N: 1 (from \(\text{NaNO}_2\))
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Right side:
- H: 1 (from \(\text{HNO}_2\))
- N: 1 (from \(\text{HNO}_2\))
- O: 2 (from \(\text{HNO}_2\)) + 4 (from \(\text{Na}_2\text{SO}_4\)) = 6
- Na: 2 (from \(\text{Na}_2\text{SO}_4\))
- S: 1 (from \(\text{Na}_2\text{SO}_4\))
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To balance, we can start adjusting coefficients:
Let's try \(a\) for \(\text{H}_2\text{SO}_4\), \(b\) for \(\text{NaNO}_2\), \(c\) for \(\text{HNO}_2\), and \(d\) for \(\text{Na}_2\text{SO}_4\):
\[ a \text{H}_2\text{SO}_4 + b \text{NaNO}_2 \rightarrow c \text{HNO}_2 + d \text{Na}_2\text{SO}_4 \]
By inspection, we see that we can balance the sodium and hydrogen properly.
After testing values, the balanced equation would be:
\[ 1 \text{H}_2\text{SO}_4 + 2 \text{NaNO}_2 \rightarrow 2 \text{HNO}_2 + 1 \text{Na}_2\text{SO}_4 \]
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Therefore, the coefficients are:
- \(\text{H}_2\text{SO}_4\): 1
- \(\text{NaNO}_2\): 2
- \(\text{HNO}_2\): 2
- \(\text{Na}_2\text{SO}_4\): 1
Thus, the balanced equation is:
\[ \boxed{1} \text{H}_2\text{SO}_4 + \boxed{2} \text{NaNO}_2 \rightarrow \boxed{2} \text{HNO}_2 + \boxed{1} \text{Na}_2\text{SO}_4 \]