Answer :
To solve this problem, let's analyze the intermediate chemical equations and determine the necessary alterations to obtain the final balanced equation.
The given intermediate chemical equations are:
1. [tex]\( 2 Na (s) + Cl_2 (g) \rightarrow 2 NaCl (s) \)[/tex]
2. [tex]\( 2 Na_2 O (s) \rightarrow 4 Na (s) + O_2 (g) \)[/tex]
We are looking for the products [tex]\( NaCl \)[/tex] and [tex]\( O_2 \)[/tex] formed from the reaction between [tex]\( Na_2 O \)[/tex] and [tex]\( Cl_2 \)[/tex].
First, let’s balance the equations individually and combine them.
1. The reaction [tex]\( 2 Na (s) + Cl_2 (g) \rightarrow 2 NaCl (s) \)[/tex] shows that 2 sodium atoms react with 1 molecule of chlorine to produce 2 molecules of sodium chloride.
2. The reaction [tex]\( 2 Na_2 O (s) \rightarrow 4 Na (s) + O_2 (g) \)[/tex] indicates that 2 formula units of sodium oxide decompose into 4 sodium atoms and 1 molecule of oxygen gas.
To combine these equations properly, we need to ensure that the number of sodium atoms is balanced.
Since we need 8 Na atoms (from 4 [tex]\( Na_2 O \)[/tex]) to react with chlorine:
- Multiply the first equation by 2 to balance the Na atoms with the second equation:
[tex]\[ 2 \times [ 2 Na (s) + Cl_2 (g) \rightarrow 2 NaCl (s) ] \][/tex]
This becomes:
[tex]\[ 4 Na (s) + 2 Cl_2 (g) \rightarrow 4 NaCl (s) \][/tex]
- Multiply the second equation by 2 to maintain consistency in the number of sodium atoms:
[tex]\[ 2 \times [ 2 Na_2 O (s) \rightarrow 4 Na (s) + O_2 (g) ] \][/tex]
This becomes:
[tex]\[ 4 Na_2 O (s) \rightarrow 8 Na (s) + 2 O_2 (g) \][/tex]
Now, we can combine these balanced equations:
1. [tex]\( 4 Na_2 O (s) \rightarrow 8 Na (s) + 2 O_2 (g) \)[/tex]
2. [tex]\( 4 Na (s) + 2 Cl_2 (g) \rightarrow 4 NaCl (s) \)[/tex]
Combining these adjusted equations gives us:
[tex]\[ 4 Na_2 O (s) + 2 Cl_2 (g) \rightarrow 4 NaCl (s) + 2 O_2 (g) \][/tex]
Therefore, the correct alterations needed to combine these equations effectively are:
- Multiply the second equation by 2.
- Multiply the first equation by 2.
The correct answer is:
1. Multiply the second equation by 2.
2. Multiply the first equation by 2.
The given intermediate chemical equations are:
1. [tex]\( 2 Na (s) + Cl_2 (g) \rightarrow 2 NaCl (s) \)[/tex]
2. [tex]\( 2 Na_2 O (s) \rightarrow 4 Na (s) + O_2 (g) \)[/tex]
We are looking for the products [tex]\( NaCl \)[/tex] and [tex]\( O_2 \)[/tex] formed from the reaction between [tex]\( Na_2 O \)[/tex] and [tex]\( Cl_2 \)[/tex].
First, let’s balance the equations individually and combine them.
1. The reaction [tex]\( 2 Na (s) + Cl_2 (g) \rightarrow 2 NaCl (s) \)[/tex] shows that 2 sodium atoms react with 1 molecule of chlorine to produce 2 molecules of sodium chloride.
2. The reaction [tex]\( 2 Na_2 O (s) \rightarrow 4 Na (s) + O_2 (g) \)[/tex] indicates that 2 formula units of sodium oxide decompose into 4 sodium atoms and 1 molecule of oxygen gas.
To combine these equations properly, we need to ensure that the number of sodium atoms is balanced.
Since we need 8 Na atoms (from 4 [tex]\( Na_2 O \)[/tex]) to react with chlorine:
- Multiply the first equation by 2 to balance the Na atoms with the second equation:
[tex]\[ 2 \times [ 2 Na (s) + Cl_2 (g) \rightarrow 2 NaCl (s) ] \][/tex]
This becomes:
[tex]\[ 4 Na (s) + 2 Cl_2 (g) \rightarrow 4 NaCl (s) \][/tex]
- Multiply the second equation by 2 to maintain consistency in the number of sodium atoms:
[tex]\[ 2 \times [ 2 Na_2 O (s) \rightarrow 4 Na (s) + O_2 (g) ] \][/tex]
This becomes:
[tex]\[ 4 Na_2 O (s) \rightarrow 8 Na (s) + 2 O_2 (g) \][/tex]
Now, we can combine these balanced equations:
1. [tex]\( 4 Na_2 O (s) \rightarrow 8 Na (s) + 2 O_2 (g) \)[/tex]
2. [tex]\( 4 Na (s) + 2 Cl_2 (g) \rightarrow 4 NaCl (s) \)[/tex]
Combining these adjusted equations gives us:
[tex]\[ 4 Na_2 O (s) + 2 Cl_2 (g) \rightarrow 4 NaCl (s) + 2 O_2 (g) \][/tex]
Therefore, the correct alterations needed to combine these equations effectively are:
- Multiply the second equation by 2.
- Multiply the first equation by 2.
The correct answer is:
1. Multiply the second equation by 2.
2. Multiply the first equation by 2.