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 ?
B. multiplying the first equation by 2.
C. multiplying the first equation by (1/2).
D. multiplying the second equation by (1/4).
Answer (1)
To eliminate N a ( s ) from the intermediate equations, determine the correct multiplier for the second equation.
Multiply the second equation by x : 2 x N a 2 O ( s ) → 4 x N a ( s ) + x O 2 ( g ) .
Set the coefficients of N a ( s ) to cancel each other out: 4 x = 2 .
Solve for x to find the multiplier: x = 2 1 . The final answer is multiplying the second equation by 2 1 .
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 multiplier for the second equation such that when we add the two equations, the N a ( s ) terms cancel out, leaving N a 2 O and C l 2 as reactants and N a Cl and O 2 as products.
Setting up the Equations Let's denote the multiplier for the second equation as x . Multiplying the second equation by x , we get:
2 x N a 2 O ( s ) → 4 x N a ( s ) + x O 2 ( g )
The first equation remains:
2 N a ( s ) + C l 2 ( g ) → 2 N a Cl ( s )
Finding the Multiplier To eliminate N a ( s ) , the coefficients of N a ( s ) in both equations must be equal in magnitude but opposite in sign. In the first equation, the coefficient of N a ( s ) is 2. In the modified second equation, the coefficient of N a ( s ) is 4 x . To cancel out the N a ( s ) terms when adding the equations, we need 4 x = − 2 ( − 1 ) , which simplifies to 4 x = 2 .
Solving for x Solving for x , we get:
x = 4 2 = 2 1
So, we need to multiply the second equation by 2 1 .
Verifying the Solution Multiplying the second equation by 2 1 , we get:
2 1 [ 2 N a 2 O ( s ) → 4 N a ( s ) + O 2 ( g )] ⟹ N a 2 O ( s ) → 2 N a ( s ) + 2 1 O 2 ( g )
Now, adding this modified equation to the first equation:
[ 2 N a ( s ) + C l 2 ( g ) → 2 N a Cl ( s )] + [ N a 2 O ( s ) → 2 N a ( s ) + 2 1 O 2 ( g )]
2 N a ( s ) + C l 2 ( g ) + N a 2 O ( s ) → 2 N a Cl ( s ) + 2 N a ( s ) + 2 1 O 2 ( g )
Cancelling the 2 N a ( s ) terms from both sides, we get the final equation:
N a 2 O ( s ) + C l 2 ( g ) → 2 N a Cl ( s ) + 2 1 O 2 ( g )
Final Answer Therefore, we need to multiply the second equation by 2 1 .
Examples
In chemical reactions, balancing equations is crucial to ensure that the number of atoms for each element is the same on both sides of the equation, adhering to the law of conservation of mass. This problem demonstrates how to manipulate intermediate equations to arrive at a balanced final equation, which is essential in fields like stoichiometry, where you calculate the amounts of reactants and products in chemical reactions. For instance, if you're producing NaCl and O 2 from N a 2 O and C l 2 , understanding these balanced equations helps determine the precise quantities of each substance needed for the reaction.
