To determine the correct balanced equation, we can analyze the original equation provided:
\[ 4 \text{NaCl} + 2 \text{SO}_2 + _ \text{H}_2\text{O} + _ \text{O}_2 \rightarrow _ \text{Na}_2\text{SO}_4 + 4 \text{HCl} \]
First, we need to find coefficients for \(\text{H}_2\text{O}\) and \(\text{O}_2\) to balance the reaction.
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Sodium (Na) in \(\text{NaCl}\) gives us 4 Na atoms on the left, thus we need:
- \(4 \text{NaCl} \rightarrow 2 \text{Na}_2\text{SO}_4\)
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Sulfur (S) in \(\text{SO}_2\) gives us 2 S atoms on the left, thus we need the same on the right:
- \(2 \text{SO}_2 \rightarrow 2 \text{Na}_2\text{SO}_4\)
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Chlorine (Cl) from \(\text{NaCl}\) gives us 4 Cl atoms, leading to:
- \(4 \text{HCl}\)
Considering hydrogen (H) and oxygen (O):
- Each \(\text{HCl}\) requires 1 \(\text{H}\), thus 4 \(\text{HCl}\) would require 4 H from 2 \(\text{H}_2\text{O}\), giving us a total of 2 \(\text{H}_2\text{O}\).
- Oxygen needs to be balanced. At this point, we check:
From the equation: \[ 2 \text{SO}_2 \] provides 4 O \[ 2 \text{H}_2\text{O} \] provides 2 O Adding these gives us 6 O, while 2 O from \(\text{O}_2\) would contribute 2 O for a total of 8 on the left.
The total amount of products needs to balance both the sodium sulfate and the HCl, giving: \[ 4 \text{Na}_2\text{SO}_4 \rightarrow 4 \text{Na}_2\text{SO}_4 + 4 \text{HCl} \]
Putting this together: 4NaCl + 2SO2 + 2H2O + O2 → 2Na2SO4 + 4HCl
Thus, the balanced equation is:
4NaCl + 2SO2 + 2H2O + O2 → 2Na2SO4 + 4HCl
The correct response is:
4NaCl + 2SO2 + 2H2O + O2 → 2Na2SO4 + 4HCl
1 answer