Question

1. ammonia can be made by reaction of water with magnesium nitride as shown by the following unbalanced equation: mg₃n₂(s) + h₂o(l) → mg(oh)₂(s) + nh₃(g) if this process is 80% efficient, what mass of ammonia can be prepared from 19.0 kg magnesium nitride? a) 2.6 kg nh3 b) 6.4 kg nh3 c) 5.1 kg nh3 d) 3.2 kg nh3 e) 15 kg nh3

Explanation:

Step1: Balance the Chemical Equation

The unbalanced equation is \( \text{Mg}_3\text{N}_2(\text{s}) + \text{H}_2\text{O}(\text{l})
ightarrow \text{Mg(OH)}_2(\text{s}) + \text{NH}_3(\text{g}) \).
To balance it, we start with Mg: there are 3 Mg in \( \text{Mg}_3\text{N}_2 \), so we put 3 in front of \( \text{Mg(OH)}_2 \).
For N: there are 2 N in \( \text{Mg}_3\text{N}_2 \), so we put 2 in front of \( \text{NH}_3 \).
For H and O: Now, \( \text{Mg(OH)}_2 \) has 2 OH⁻ groups, 3 of them give 6 O and 6 H. \( \text{NH}_3 \) has 2×3 = 6 H. So total H from water should be 6 (from \( \text{Mg(OH)}_2 \)) + 6 (from \( \text{NH}_3 \)) = 12 H, so we need 6 \( \text{H}_2\text{O} \) (since each \( \text{H}_2\text{O} \) has 2 H). Checking O: 6 \( \text{H}_2\text{O} \) has 6 O, and 3 \( \text{Mg(OH)}_2 \) has 6 O, so O is balanced.
Balanced equation: \( \text{Mg}_3\text{N}_2(\text{s}) + 6\text{H}_2\text{O}(\text{l})
ightarrow 3\text{Mg(OH)}_2(\text{s}) + 2\text{NH}_3(\text{g}) \).

Step2: Calculate Molar Masses

• Molar mass of \( \text{Mg}_3\text{N}_2 \): \( 3\times24.31 + 2\times14.01 = 72.93 + 28.02 = 100.95 \, \text{g/mol} \) (or \( 0.10095 \, \text{kg/mol} \)).
• Molar mass of \( \text{NH}_3 \): \( 14.01 + 3\times1.008 = 17.034 \, \text{g/mol} \) (or \( 0.017034 \, \text{kg/mol} \)).

Step3: Moles of \( \text{Mg}_3\text{N}_2 \)

Given mass of \( \text{Mg}_3\text{N}_2 = 19.0 \, \text{kg} \).
Moles of \( \text{Mg}_3\text{N}_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{19.0 \, \text{kg}}{0.10095 \, \text{kg/mol}} \approx 188.2 \, \text{mol} \).

Step4: Moles of \( \text{NH}_3 \) from Stoichiometry

From the balanced equation, 1 mol \( \text{Mg}_3\text{N}_2 \) produces 2 mol \( \text{NH}_3 \).
So moles of \( \text{NH}_3 \) (theoretical) = \( 2 \times 188.2 \, \text{mol} = 376.4 \, \text{mol} \).

Step5: Theoretical Mass of \( \text{NH}_3 \)

Mass (theoretical) = moles × molar mass = \( 376.4 \, \text{mol} \times 0.017034 \, \text{kg/mol} \approx 6.41 \, \text{kg} \).

Step6: Actual Mass (Considering Efficiency)

Efficiency = 80% = 0.8.
Actual mass of \( \text{NH}_3 = \text{Theoretical mass} \times \text{Efficiency} = 6.41 \, \text{kg} \times 0.8 \approx 5.13 \, \text{kg} \), which is approximately 5.1 kg (matching option C).

Answer:

C. 5.1 kg NH3