To balance the reaction:
\[ 4 \text{NaCl} + 2 \text{SO}_2 + 4 \text{H}_2\text{O} + _ \text{O}_2 \rightarrow _ \text{Na}_2\text{SO}_4 + 4 \text{HCl} \]
Let's analyze the reaction.
- The reactants include sodium chloride (NaCl), sulfur dioxide (SO2), and water (H2O). The products include sodium sulfate (Na2SO4) and hydrochloric acid (HCl).
- You have 4 NaCl, which means you will produce 4 moles of HCl since each sodium chloride will give rise to one HCl molecule.
- For the sodium sulfate (Na2SO4), you produce 2 moles, because each sodium sulfate contains 2 sodium ions (Na+).
From this, we can start balancing the oxygen:
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On the product side:
- Each Na2SO4 has 4 O from 2 Na2SO4 = 4 O
- The 4 HCl do not contribute to the oxygen count.
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Thus, we have a total of 4 O on the products side from the sodium sulfate.
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On the reactant side:
- We have 2 SO2 contributing 4 O (2×2=4).
- We have 4 H2O contributing 4 O (4×1=4).
Now, incorporating oxygen from O2: \[ \text{Total O on the reactant side} = 4 \text{(from SO}_2\text{)} + 4 \text{(from H}_2\text{O)} + _ \text{O}_2 \] We already have 8 O from the sulfites and water, so we do not need any additional O2. Thus, the first blank is filled with 0.
For the second blank (number of moles of Na2SO4), since we have:
- 2 moles of Na2SO4 produced from the 4 Na (since each Na2SO4 uses 2 Na), we fill the second blank with 2.
So filling in the values:
\[ 4 \text{NaCl} + 2 \text{SO}_2 + 4 \text{H}_2\text{O} + 0 \text{O}_2 \rightarrow 2 \text{Na}_2\text{SO}_4 + 4 \text{HCl} \]
Final answer: First blank: 0, Second blank: 2.
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