Barium nitride (Ba3N2) and water (H2O) react to produce barium hydroxide (Ba(OH)2) and ammonia (NH3)

Barium nitride (Ba3N2) and water (H2O) react to produce barium hydroxide (Ba(OH)2) and ammonia (NH3). Balance the chemical reaction

Ba3N2+H2O ? Ba(OH)2+NH3

as follows:

(a) Write down the system of linear equations for the quantities of each chemical necessary to balance the reaction. Explain briefly how you got them.

(b) Reduce the augmented matrix of the system you found in (a) to reduced row echelon form. Show at least 4 stages of the reduction process and list all of the row operations that you use to get from each stage to the next.

(c) Describe the solutions as multiples of a fundamental vector and interpret your answer in terms of the chemicals above.

The Correct Answer and Explanation is :

Let’s go step by step through the process of balancing the reaction for Barium nitride (Ba3N2) and water (H2O) reacting to produce barium hydroxide (Ba(OH)2) and ammonia (NH3).

(a) Write down the system of linear equations for the quantities of each chemical necessary to balance the reaction.

The unbalanced reaction is:

[
\text{Ba}_3\text{N}_2 + \text{H}_2\text{O} \rightarrow \text{Ba(OH)}_2 + \text{NH}_3
]

Let the coefficients of the chemicals be:

• ( a ) for Ba3N2
• ( b ) for H2O
• ( c ) for Ba(OH)2
• ( d ) for NH3

The system of linear equations is based on the conservation of each element (Ba, N, O, and H) on both sides of the equation.

1. Barium (Ba): On the left side, Ba3N2 has 3 Ba atoms. On the right side, Ba(OH)2 has 1 Ba atom per molecule, so:
   [
   3a = c
   ]
2. Nitrogen (N): On the left side, Ba3N2 has 2 N atoms. On the right side, NH3 has 1 N atom per molecule, so:
   [
   2a = d
   ]
3. Oxygen (O): On the left side, H2O has 1 O atom per molecule. On the right side, Ba(OH)2 has 2 O atoms per molecule, so:
   [
   b = 2c
   ]
4. Hydrogen (H): On the left side, H2O has 2 H atoms per molecule. On the right side, Ba(OH)2 has 2 H atoms per molecule and NH3 has 3 H atoms per molecule, so:
   [
   2b = 2c + 3d
   ]

Now we have the following system of equations:

[
\begin{aligned}
3a &= c \
2a &= d \
b &= 2c \
2b &= 2c + 3d
\end{aligned}
]

(b) Reduce the augmented matrix of the system you found in (a) to reduced row echelon form.

The augmented matrix corresponding to the system of equations is:

[
\begin{bmatrix}
3 & 0 & -1 & 0 & 0 \
2 & 0 & 0 & -1 & 0 \
0 & 1 & -2 & 0 & 0 \
0 & 2 & -2 & -3 & 0
\end{bmatrix}
]

Now, we’ll perform row operations to reduce this to reduced row echelon form (RREF). The goal is to create leading ones and eliminate other non-zero entries in each column.

1. Row operation 1 (R2 → R2 – (2/3)R1):
   [
   \begin{bmatrix}
   3 & 0 & -1 & 0 & 0 \
   0 & 0 & 2/3 & -1 & 0 \
   0 & 1 & -2 & 0 & 0 \
   0 & 2 & -2 & -3 & 0
   \end{bmatrix}
   ]
2. Row operation 2 (R3 → R3 + (2/3)R1):
   [
   \begin{bmatrix}
   3 & 0 & -1 & 0 & 0 \
   0 & 0 & 2/3 & -1 & 0 \
   0 & 1 & -2/3 & 0 & 0 \
   0 & 2 & -2 & -3 & 0
   \end{bmatrix}
   ]
3. Row operation 3 (R4 → R4 – 2R3):
   [
   \begin{bmatrix}
   3 & 0 & -1 & 0 & 0 \
   0 & 0 & 2/3 & -1 & 0 \
   0 & 1 & -2/3 & 0 & 0 \
   0 & 0 & 0 & -3 & 0
   \end{bmatrix}
   ]
4. Row operation 4 (R4 → R4 ÷ -3):
   [
   \begin{bmatrix}
   3 & 0 & -1 & 0 & 0 \
   0 & 0 & 2/3 & -1 & 0 \
   0 & 1 & -2/3 & 0 & 0 \
   0 & 0 & 0 & 1 & 0
   \end{bmatrix}
   ]

(c) Describe the solutions as multiples of a fundamental vector and interpret your answer in terms of the chemicals above.

The final matrix represents the system with the following solution:

[
\begin{aligned}
a &= c \
b &= 2c \
d &= 2a
\end{aligned}
]

Let ( c = t ) (where ( t ) is any real number). Then:

[
a = t, \quad b = 2t, \quad d = 2t
]

Thus, the solution is in the form of a multiple of the vector ( (1, 2, 1, 2) ), where each element corresponds to the coefficients of Ba3N2, H2O, Ba(OH)2, and NH3, respectively.

This means that the balanced chemical equation is:

[
\text{Ba}_3\text{N}_2 + 3\text{H}_2\text{O} \rightarrow 3\text{Ba(OH)}_2 + 2\text{NH}_3
]

This solution implies that for every 1 molecule of Ba3N2, 3 molecules of water react to produce 3 molecules of barium hydroxide and 2 molecules of ammonia.

Let me generate an image of the process.

Here is the diagram illustrating the step-by-step reduction of the system of linear equations, showing how the matrix is transformed to reduced row echelon form. Each stage of the process and the row operations are clearly labeled to help visualize how we reach the solution. Let me know if you’d like further explanation or any adjustments!