Consider the following intermediate chemical equations:
[tex]
\begin{array}{l}
2 Na(s)+Cl(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)
To yield the desired overall reaction from the given intermediate equations, you need to multiply the first equation by 2. This allows for the production of the required amount of sodium chloride and oxygen. Therefore, the answer is option B: multiplying the first equation by 2.
;
Balance the target overall reaction: 2 N a 2 O ( s ) + 2 C l 2 ( g ) → 4 N a Cl ( s ) + O 2 ( g ) .
Multiply the first intermediate equation by 2: 4 N a ( s ) + 2 C l 2 ( g ) → 4 N a Cl ( s ) .
Keep the second intermediate equation as is: 2 N a 2 O ( s ) → 4 N a ( s ) + O 2 ( g ) .
Adding the two equations gives the balanced overall reaction, so the first equation is multiplied by 2. multiplying the first equation by 2 .
Explanation
Understanding the Problem We are given two intermediate chemical equations and need to determine the correct multipliers for these equations so that when added, they yield a final equation with N a Cl and O 2 as products formed from N a 2 O and C l 2 .
Analyzing the Equations The given intermediate equations are:
2 N a ( s ) + C l 2 ( g ) i g h t ha r p oo n u p 2 N a Cl ( s )
2 N a 2 O ( s ) i g h t ha r p oo n u p 4 N a ( s ) + O 2 ( g )
We want to find the multipliers for these equations such that when added, they result in the reaction:
2 N a 2 O ( s ) + 2 C l 2 ( g ) i g h t ha r p oo n u p 4 N a Cl ( s ) + O 2 ( g )
Multiplying the First Equation To eliminate N a ( s ) , we need to manipulate the given equations. Observe that in the second equation, 4 N a ( s ) is produced. To cancel this out, we need 4 N a ( s ) to be consumed in the first equation. Thus, we multiply the first equation by 2:
2 { 2 N a ( s ) + C l 2 ( g ) ⇀ 2 N a Cl ( s )} ⟹ 4 N a ( s ) + 2 C l 2 ( g ) ⇀ 4 N a Cl ( s )
Keeping the Second Equation Now, we keep the second equation as is:
2 N a 2 O ( s ) ⇀ 4 N a ( s ) + O 2 ( g )
Adding the Equations Adding the modified first equation and the second equation, we get:
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 )
Simplifying by cancelling out 4 N a ( s ) from both sides, we obtain:
2 N a 2 O ( s ) + 2 C l 2 ( g ) ⇀ 4 N a Cl ( s ) + O 2 ( g )
This matches the balanced target equation.
Conclusion Therefore, we only needed to multiply the first equation by 2.
Examples
In chemical engineering, balancing equations is crucial for designing industrial processes. For example, if you're designing a process to produce sodium chloride ( N a Cl ) and oxygen ( O 2 ) from sodium oxide ( N a 2 O ) and chlorine ( C l 2 ), you need to understand the stoichiometry of the reaction. By correctly balancing the intermediate equations, you can determine the exact amounts of reactants needed and products formed, ensuring efficient and safe operation of the chemical plant. This ensures minimal waste and maximal yield, which is both economically and environmentally beneficial.
