To find the overall equation for the reaction that produces NaCl and O₂ from Na₂O and Cl₂, we need to follow a systematic approach.
Firstly, we note the starting materials (reactants) given in the problem statement:
- Na₂O (s)
- Cl₂ (g)
And the products we're aiming to produce:
- NaCl (s)
- O₂ (g)
Next, we look at the coefficients from the original individual reactions:
1. \( 2 Na (s) + Cl_2 (g) \rightarrow 2 NaCl (s) \)
2. \( 4 Na (s) + O_2 (g) \rightarrow 2 Na_2O (s) \)
In these reactions, sodium (Na) reacts either with chlorine (Cl₂) to form sodium chloride (NaCl) or with oxygen (O₂) to form sodium oxide (Na₂O). To combine these to form NaCl and O₂ directly from Na₂O and Cl₂, we need to balance the atoms on both the reactant and product sides.
From the equations:
- Reaction 1 shows that 2 Na reacts with 1 Cl₂ to form 2 NaCl.
- Reaction 2 shows that 4 Na reacts with 1 O₂ to form 2 Na₂O.
To go in the reverse direction of Reaction 2 for forming Na₂O and O₂:
[tex]\[ 2 Na_2O (s) \rightarrow 4 Na (s) + O_2 (g) \][/tex]
Now, we can combine these with Reaction 1 (in the forward direction):
[tex]\[ 2 Na (s) + Cl_2 (g) \rightarrow 2 NaCl (s) \][/tex]
To balance the sodium (Na) atoms, since we need 4 Na atoms from the reverse of Reaction 2:
[tex]\[ 4 Na (s) + 2 Cl_2 (g) \rightarrow 4 NaCl (s) \][/tex]
Finally, we combine these:
[tex]\[ 2 Na_2O (s) + 2 Cl_2 (g) \rightarrow 4 NaCl (s) + O_2 (g) \][/tex]
Hence, the overall equation is:
[tex]\[ 2 Na_2O (s) + 2 Cl_2 (g) \rightarrow 4 NaCl (s) + O_2 (g) \][/tex]
Therefore, the correct answer is:
[tex]\[ \boxed{2 Na_2O (s) + 2 Cl_2 (g) \rightarrow 4 NaCl (s) + O_2 (g)} \][/tex]
