Answer :
To address the problem of balancing the given chemical reactions to form the final chemical equation, follow these steps:
1. Start with the given intermediate equations:
[tex]\[ 2 Na (s) + Cl_2 (g) \rightarrow 2 NaCl (s) \][/tex]
[tex]\[ 2 Na_2O (s) \rightarrow 4 Na (s) + O_2 (g) \][/tex]
2. Analyze the final desired reaction, where [tex]\(Na_2O\)[/tex] and [tex]\(Cl_2\)[/tex] react to form [tex]\(NaCl\)[/tex] and [tex]\(O_2\)[/tex]. Start by examining the number of atoms involved:
[tex]\[ \text{Final equation: } Na_2O + Cl_2 \rightarrow NaCl + O_2 \][/tex]
3. To achieve the final equation, we need to ensure the number of atoms on both sides of the reaction is balanced. Focus on the species involved in the intermediate reactions:
- From the first equation: 2 Na (s) + Cl_2 (g) \rightarrow 2 NaCl (s)
- From the second equation: 2 Na_2O (s) \rightarrow 4 Na (s) + O_2 (g)
4. Notice that the second equation produces 4 Na (s) while the first equation requires 2 Na (s). To balance it correctly, observe that the numbers of Na atoms produced and used need to match in both intermediate and final equations.
5. If we multiply the first equation by 2:
[tex]\[ 2 \times [2 Na (s) + Cl_2 (g) \rightarrow 2 NaCl (s)] \][/tex]
It becomes:
[tex]\[ 4 Na (s) + 2 Cl_2 (g) \rightarrow 4 NaCl (s) \][/tex]
6. The second equation remains as is:
[tex]\[ 2 Na_2O (s) \rightarrow 4 Na (s) + O_2 (g) \][/tex]
7. By combining these equations, we add the intermediate reactions to form the final balanced equation:
[tex]\[ 2 Na_2O (s) + 4 Na (s) + 2 Cl_2 (g) \rightarrow 4 Na (s) + 4 NaCl (s) + O_2 (g) \][/tex]
8. Cancel out the 4 Na (s) present on both sides to simplify the equation:
[tex]\[ 2 Na_2O (s) + 2 Cl_2 (g) \rightarrow 4 NaCl (s) + O_2 (g) \][/tex]
9. Therefore, we needed to multiply the first equation by 2 in order to ensure the final equation was balanced properly.
Answer: We need to multiply the first equation by 2.
1. Start with the given intermediate equations:
[tex]\[ 2 Na (s) + Cl_2 (g) \rightarrow 2 NaCl (s) \][/tex]
[tex]\[ 2 Na_2O (s) \rightarrow 4 Na (s) + O_2 (g) \][/tex]
2. Analyze the final desired reaction, where [tex]\(Na_2O\)[/tex] and [tex]\(Cl_2\)[/tex] react to form [tex]\(NaCl\)[/tex] and [tex]\(O_2\)[/tex]. Start by examining the number of atoms involved:
[tex]\[ \text{Final equation: } Na_2O + Cl_2 \rightarrow NaCl + O_2 \][/tex]
3. To achieve the final equation, we need to ensure the number of atoms on both sides of the reaction is balanced. Focus on the species involved in the intermediate reactions:
- From the first equation: 2 Na (s) + Cl_2 (g) \rightarrow 2 NaCl (s)
- From the second equation: 2 Na_2O (s) \rightarrow 4 Na (s) + O_2 (g)
4. Notice that the second equation produces 4 Na (s) while the first equation requires 2 Na (s). To balance it correctly, observe that the numbers of Na atoms produced and used need to match in both intermediate and final equations.
5. If we multiply the first equation by 2:
[tex]\[ 2 \times [2 Na (s) + Cl_2 (g) \rightarrow 2 NaCl (s)] \][/tex]
It becomes:
[tex]\[ 4 Na (s) + 2 Cl_2 (g) \rightarrow 4 NaCl (s) \][/tex]
6. The second equation remains as is:
[tex]\[ 2 Na_2O (s) \rightarrow 4 Na (s) + O_2 (g) \][/tex]
7. By combining these equations, we add the intermediate reactions to form the final balanced equation:
[tex]\[ 2 Na_2O (s) + 4 Na (s) + 2 Cl_2 (g) \rightarrow 4 Na (s) + 4 NaCl (s) + O_2 (g) \][/tex]
8. Cancel out the 4 Na (s) present on both sides to simplify the equation:
[tex]\[ 2 Na_2O (s) + 2 Cl_2 (g) \rightarrow 4 NaCl (s) + O_2 (g) \][/tex]
9. Therefore, we needed to multiply the first equation by 2 in order to ensure the final equation was balanced properly.
Answer: We need to multiply the first equation by 2.
