Question 40

Question

Match the missing coefficient to the correct number of molecules so that the equation adheres to the conservation of mass.

(X)N2 + (Y)H2 → (Z)NH3

Column A
1.
X:
X
2.
Y:
Y
3.
Z:
Z
Column B
a.3
b.1
c.2

Answer 1

To balance the chemical equation \( (X)N_2 + (Y)H_2 → (Z)NH_3 \) while adhering to the conservation of mass, we can analyze the equation step by step.

1. Nitrogen (N) atoms: On the left side, each \( N_2 \) molecule contributes 2 nitrogen atoms. Let \( X \) be the coefficient for \( N_2 \). Then, the total number of nitrogen atoms is \( 2X \).
2. Hydrogen (H) atoms: On the left side, each \( H_2 \) molecule contributes 2 hydrogen atoms. Let \( Y \) be the coefficient for \( H_2 \). Thus, the total number of hydrogen atoms is \( 2Y \).
3. Ammonia (NH3) production: Each \( NH_3 \) molecule has 1 nitrogen and 3 hydrogen atoms. Let \( Z \) be the coefficient for \( NH_3 \). Therefore, the total number of nitrogen and hydrogen atoms in the products is \( Z \) for nitrogen and \( 3Z \) for hydrogen.

Setting up the equations based on the conservation of mass:

• For nitrogen: \( 2X = Z \)
• For hydrogen: \( 2Y = 3Z \)

From these equations, we can derive values for \( X \), \( Y \), and \( Z \).

1. Let’s choose \( Z = 2 \) (it’s a simple whole number to start).
   • From \( 2X = Z \) → \( 2X = 2 \) → \( X = 1 \).
   • From \( 2Y = 3Z \) → \( 2Y = 3(2) \) → \( 2Y = 6 \) → \( Y = 3 \).

This gives us the following coefficients:

• \( X = 1 \)
• \( Y = 3 \)
• \( Z = 2 \)

Now, matching these values with the options provided in Columns A and B:

Column A

1. \( X: 1 \) → b
2. \( Y: 3 \) → a
3. \( Z: 2 \) → c

So the final matches are:

• \( X \) matches with \( b (1) \)
• \( Y \) matches with \( a (3) \)
• \( Z \) matches with \( c (2) \)

Thus, the answers for the coefficients are:

• \( X: 1 \)
• \( Y: 3 \)
• \( Z: 2 \)