Answer :
To determine the correct balanced equation, we need to ensure that the number of atoms of each element on the left side of the equation equals the number on the right side.
Let's start with the unbalanced equation:
[tex]\[ NH_3 + O_2 \rightarrow NO + H_2O \][/tex]
We will use coefficients to balance the number of atoms of each element on both sides. Let's denote the coefficients as follows:
[tex]\[ a \cdot NH_3 + b \cdot O_2 \rightarrow c \cdot NO + d \cdot H_2O \][/tex]
We need to balance the number of each type of atom (N, H, and O) on both sides of the equation.
1. Balance Nitrogen (N) atoms:
On the left side, we have [tex]\(a \cdot NH_3\)[/tex], which contains [tex]\(a\)[/tex] nitrogen atoms.
On the right side, we have [tex]\(c \cdot NO\)[/tex], which contains [tex]\(c\)[/tex] nitrogen atoms.
Therefore:
[tex]\[ a = c \][/tex]
2. Balance Hydrogen (H) atoms:
On the left side, we have [tex]\(a \cdot NH_3\)[/tex], which contains [tex]\(3a\)[/tex] hydrogen atoms.
On the right side, we have [tex]\(d \cdot H_2O\)[/tex], which contains [tex]\(2d\)[/tex] hydrogen atoms.
Therefore:
[tex]\[ 3a = 2d \][/tex]
3. Balance Oxygen (O) atoms:
On the left side, we have [tex]\(b \cdot O_2\)[/tex], which contains [tex]\(2b\)[/tex] oxygen atoms.
On the right side, we have [tex]\(c \cdot NO\)[/tex] and [tex]\(d \cdot H_2O\)[/tex], which together contain [tex]\(c + d\)[/tex] oxygen atoms.
Therefore:
[tex]\[ 2b = c + d \][/tex]
Now let's solve these equations step-by-step.
From the first equation [tex]\(a = c\)[/tex].
From the second equation [tex]\(3a = 2d\)[/tex], we can solve for [tex]\(d\)[/tex]:
[tex]\[ d = \frac{3a}{2} \][/tex]
Substitute [tex]\(d = \frac{3a}{2}\)[/tex] into the third equation [tex]\(2b = c + d\)[/tex]:
[tex]\[ 2b = a + \frac{3a}{2} \][/tex]
Simplify:
[tex]\[ 2b = \frac{2a + 3a}{2} \][/tex]
[tex]\[ 2b = \frac{5a}{2} \][/tex]
Multiply both sides by 2:
[tex]\[ 4b = 5a \][/tex]
Solve for [tex]\(b\)[/tex]:
[tex]\[ b = \frac{5a}{4} \][/tex]
We can choose the smallest integer value for [tex]\(a\)[/tex] to keep [tex]\(b\)[/tex], [tex]\(c\)[/tex], and [tex]\(d\)[/tex] as integers. If we choose [tex]\(a = 4\)[/tex]:
Then:
[tex]\[ c = 4 \][/tex]
Substitute [tex]\(a = 4\)[/tex] into [tex]\(d = \frac{3a}{2}\)[/tex]:
[tex]\[ d = \frac{3 \cdot 4}{2} = 6 \][/tex]
Substitute [tex]\(a = 4\)[/tex] into [tex]\(b = \frac{5a}{4}\)[/tex]:
[tex]\[ b = \frac{5 \cdot 4}{4} = 5 \][/tex]
So the coefficients are:
[tex]\[ a = 4, \quad b = 5, \quad c = 4, \quad d = 6 \][/tex]
Substituting these back into the equation, we get:
[tex]\[ 4NH_3 + 5O_2 \rightarrow 4NO + 6H_2O \][/tex]
Therefore, the correct balanced equation is:
[tex]\[ 4NH_3 + 5O_2 \rightarrow 4NO + 6H_2O \][/tex]
The correct answer is option C.
Let's start with the unbalanced equation:
[tex]\[ NH_3 + O_2 \rightarrow NO + H_2O \][/tex]
We will use coefficients to balance the number of atoms of each element on both sides. Let's denote the coefficients as follows:
[tex]\[ a \cdot NH_3 + b \cdot O_2 \rightarrow c \cdot NO + d \cdot H_2O \][/tex]
We need to balance the number of each type of atom (N, H, and O) on both sides of the equation.
1. Balance Nitrogen (N) atoms:
On the left side, we have [tex]\(a \cdot NH_3\)[/tex], which contains [tex]\(a\)[/tex] nitrogen atoms.
On the right side, we have [tex]\(c \cdot NO\)[/tex], which contains [tex]\(c\)[/tex] nitrogen atoms.
Therefore:
[tex]\[ a = c \][/tex]
2. Balance Hydrogen (H) atoms:
On the left side, we have [tex]\(a \cdot NH_3\)[/tex], which contains [tex]\(3a\)[/tex] hydrogen atoms.
On the right side, we have [tex]\(d \cdot H_2O\)[/tex], which contains [tex]\(2d\)[/tex] hydrogen atoms.
Therefore:
[tex]\[ 3a = 2d \][/tex]
3. Balance Oxygen (O) atoms:
On the left side, we have [tex]\(b \cdot O_2\)[/tex], which contains [tex]\(2b\)[/tex] oxygen atoms.
On the right side, we have [tex]\(c \cdot NO\)[/tex] and [tex]\(d \cdot H_2O\)[/tex], which together contain [tex]\(c + d\)[/tex] oxygen atoms.
Therefore:
[tex]\[ 2b = c + d \][/tex]
Now let's solve these equations step-by-step.
From the first equation [tex]\(a = c\)[/tex].
From the second equation [tex]\(3a = 2d\)[/tex], we can solve for [tex]\(d\)[/tex]:
[tex]\[ d = \frac{3a}{2} \][/tex]
Substitute [tex]\(d = \frac{3a}{2}\)[/tex] into the third equation [tex]\(2b = c + d\)[/tex]:
[tex]\[ 2b = a + \frac{3a}{2} \][/tex]
Simplify:
[tex]\[ 2b = \frac{2a + 3a}{2} \][/tex]
[tex]\[ 2b = \frac{5a}{2} \][/tex]
Multiply both sides by 2:
[tex]\[ 4b = 5a \][/tex]
Solve for [tex]\(b\)[/tex]:
[tex]\[ b = \frac{5a}{4} \][/tex]
We can choose the smallest integer value for [tex]\(a\)[/tex] to keep [tex]\(b\)[/tex], [tex]\(c\)[/tex], and [tex]\(d\)[/tex] as integers. If we choose [tex]\(a = 4\)[/tex]:
Then:
[tex]\[ c = 4 \][/tex]
Substitute [tex]\(a = 4\)[/tex] into [tex]\(d = \frac{3a}{2}\)[/tex]:
[tex]\[ d = \frac{3 \cdot 4}{2} = 6 \][/tex]
Substitute [tex]\(a = 4\)[/tex] into [tex]\(b = \frac{5a}{4}\)[/tex]:
[tex]\[ b = \frac{5 \cdot 4}{4} = 5 \][/tex]
So the coefficients are:
[tex]\[ a = 4, \quad b = 5, \quad c = 4, \quad d = 6 \][/tex]
Substituting these back into the equation, we get:
[tex]\[ 4NH_3 + 5O_2 \rightarrow 4NO + 6H_2O \][/tex]
Therefore, the correct balanced equation is:
[tex]\[ 4NH_3 + 5O_2 \rightarrow 4NO + 6H_2O \][/tex]
The correct answer is option C.
