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)
To balance the given chemical equations, we analyze the target reaction and the provided intermediate reactions. By setting up a system of equations based on the coefficients of the reactants and products, we find that multiplying the first equation by 2 and keeping the second equation as is allows us to obtain the balanced target equation. Therefore, the correct answer is multiplying the first equation by 2.
Explanation
Analyzing the Problem We are given two intermediate chemical equations:
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 final chemical equation where N a Cl and O 2 are the products formed from the reaction between N a 2 O and C l 2 . Our goal is to determine the correct multipliers for the intermediate equations so that when they are added, the result is the desired reaction.
Writing the Balanced Target Equation Let's write the target equation:
N a 2 O ( s ) + C l 2 ( g ) i g h t ha r p oo n u pN a Cl ( s ) + O 2 ( g )
However, to balance the equation correctly, we need to consider the stoichiometry. The balanced target equation is:
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 )
Modifying and Adding the Equations Now, let's multiply the first equation by a factor x and the second equation by a factor y . The modified equations are:
2 x N a ( s ) + x C l 2 ( g ) i g h t ha r p oo n u p 2 x N a Cl ( s )
2 y N a 2 O ( s ) i g h t ha r p oo n u p 4 y N a ( s ) + y O 2 ( g )
Adding the two modified equations, we get:
2 x N a ( s ) + x C l 2 ( g ) + 2 y N a 2 O ( s ) i g h t ha r p oo n u p 2 x N a Cl ( s ) + 4 y N a ( s ) + y O 2 ( g )
Eliminating Sodium Rearranging the terms to match the target equation:
2 y N a 2 O ( s ) + x C l 2 ( g ) + ( 2 x − 4 y ) N a ( s ) i g h t ha r p oo n u p 2 x N a Cl ( s ) + y O 2 ( g )
We want to eliminate N a ( s ) from the reactants. Thus, 2 x − 4 y = 0 , which simplifies to x = 2 y .
Comparing Coefficients Comparing the coefficients with the balanced target equation 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 ) , we have:
Coefficient of N a 2 O : 2 y = 2 , so y = 1 .
Coefficient of C l 2 : x = 2 .
Coefficient of N a Cl : 2 x = 4 , so x = 2 .
Coefficient of O 2 : y = 1 .
Determining the Correct Multiplier Therefore, we need to multiply the first equation by 2 and the second equation by 1. The question asks which of the given options is correct. The options are:
multiplying the second equation by 2
multiplying the first equation by 2
multiplying the first equation by (1/2)
multiplying the second equation by (1/4)
The correct option is multiplying the first equation by 2.
Examples
In chemical engineering, balancing equations is crucial for designing chemical processes. For instance, in the production of sodium chloride (table salt) from sodium oxide and chlorine gas, understanding the stoichiometry and balancing the equations ensures that the reactants are used efficiently and the desired products are obtained in the correct proportions. This principle is also applicable in environmental science, where balancing chemical equations helps in understanding and mitigating pollution by ensuring that reactions are complete and byproducts are minimized.
To balance the given intermediate chemical equations and achieve the desired products, you should multiply the first equation by 2. This adjustment ensures the reactions match the desired stoichiometry. Therefore, the correct answer is option B.
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