Consider the following intermediate chemical equations:
[tex]$\begin{array}{l}
2 Na(s)+Cl 2(g) \rightarrow 2 NaCl(s) \
2 Na_2 O(s) \rightarrow 4 Na(s)+O_2(g)
\end{array}$[/tex]
In the final chemical equation, NaCl and [tex]$O _2$[/tex] are the products that are formed through the reaction between [tex]$Na _2 O$[/tex] and [tex]$Cl _2$[/tex]. Before you can add these intermediate chemical equations, you need to alter them by
A. multiplying the second equation by 2.
B. multiplying the first equation by 2.
C. multiplying the first equation by (1/2).
D. multiplying the second equation by (1/4).
Answer (2)
In order to combine the two intermediate chemical equations correctly, we need to multiply the first equation by 2. This ensures that sodium in the reactants cancels out when added to the products. Therefore, the correct choice is option B: multiplying the first equation by 2.
;
Determine multipliers m 1 and m 2 for the two equations.
Set up the equation 2 m 1 = 4 m 2 to ensure N a ( s ) cancels out.
Solve for m 1 and m 2 , finding m 1 = 2 and m 2 = 1 .
Conclude that the first equation needs to be multiplied by 2. The answer is multiplying the first equation by 2.
Explanation
Understanding the Problem We are given two intermediate chemical equations:
2 N a ( s ) + C l 2 ( g ) → 2 N a Cl ( s )
2 N a 2 O ( s ) → 4 N a ( s ) + O 2 ( g )
We want to find the correct multipliers for these equations so that when we add them together, the intermediate N a ( s ) cancels out, and the final equation has N a Cl and O 2 as products and N a 2 O and C l 2 as reactants.
Applying Multipliers Let's denote the multipliers for the first and second equations as m 1 and m 2 , respectively.
Multiplying the first equation by m 1 gives:
2 m 1 N a ( s ) + m 1 C l 2 ( g ) → 2 m 1 N a Cl ( s )
Multiplying the second equation by m 2 gives:
2 m 2 N a 2 O ( s ) → 4 m 2 N a ( s ) + m 2 O 2 ( g )
Balancing Sodium Adding the two modified equations, we get:
( 2 m 1 N a ( s ) + m 1 C l 2 ( g ) + 2 m 2 N a 2 O ( s )) → ( 2 m 1 N a Cl ( s ) + 4 m 2 N a ( s ) + m 2 O 2 ( g ))
We want the N a ( s ) to cancel out. This means that the amount of N a ( s ) on the reactant side should be equal to the amount of N a ( s ) that turns into product. So, we need:
2 m 1 = 4 m 2
Which simplifies to:
m 1 = 2 m 2
Verifying the Solution If we choose m 2 = 1 , then m 1 = 2 . This means we multiply the first equation by 2 and the second equation by 1.
Let's check if this works:
Multiplying the first equation by 2 gives:
4 N a ( s ) + 2 C l 2 ( g ) → 4 N a Cl ( s )
Multiplying the second equation by 1 gives:
2 N a 2 O ( s ) → 4 N a ( s ) + O 2 ( g )
Adding these gives:
4 N a ( s ) + 2 C l 2 ( g ) + 2 N a 2 O ( s ) → 4 N a Cl ( s ) + 4 N a ( s ) + O 2 ( g )
The N a ( s ) cancels out, leaving:
2 N a 2 O ( s ) + 2 C l 2 ( g ) → 4 N a Cl ( s ) + O 2 ( g )
This matches the desired form.
Final Answer Therefore, we need to multiply the first equation by 2 and the second equation by 1. The question asks what we need to alter the equations by. The correct answer is multiplying the first equation by 2.
Examples
In chemical engineering, balancing equations is crucial for designing industrial processes. For instance, when synthesizing a new material, chemists must ensure that the input reactants are in the correct stoichiometric ratios to maximize product yield and minimize waste. This problem demonstrates how to adjust intermediate reactions to achieve a desired overall reaction, ensuring that no reactants are left unused and no unwanted byproducts are formed. This is also applicable in environmental science, where balancing chemical equations helps in understanding and mitigating pollution by ensuring complete reactions and preventing the release of harmful substances.
