Question

balance the following reaction between magnesium nitride and water. if there is no coefficient needed to balance the equation, place a \1\ as that answer. choose... ▼ mg₃n₂(s) + choose... ▼ h₂o(l) → choose... ▼ nh₃(g) + choose... ▼ mg(oh)₂(s)

Explanation:

Step1: Balance Mg

On the left, we have \( \text{Mg}_3\text{N}_2 \), so there are 3 Mg atoms. On the right, \( \text{Mg(OH)}_2 \) has 1 Mg atom. So we need a coefficient of 3 for \( \text{Mg(OH)}_2 \). Now the equation is \( \text{Mg}_3\text{N}_2(s) + \text{H}_2\text{O}(l)
ightarrow \text{NH}_3(g) + 3\text{Mg(OH)}_2(s) \).

Step2: Balance N

On the left, \( \text{Mg}_3\text{N}_2 \) has 2 N atoms. On the right, \( \text{NH}_3 \) has 1 N atom. So we need a coefficient of 2 for \( \text{NH}_3 \). Now the equation is \( \text{Mg}_3\text{N}_2(s) + \text{H}_2\text{O}(l)
ightarrow 2\text{NH}_3(g) + 3\text{Mg(OH)}_2(s) \).

Step3: Balance H and O

Let's count H and O. On the right, \( 2\text{NH}_3 \) has \( 2\times3 = 6 \) H atoms, and \( 3\text{Mg(OH)}_2 \) has \( 3\times2 = 6 \) H atoms from OH and \( 3\times2 = 6 \) O atoms. Wait, no, \( 3\text{Mg(OH)}_2 \) has \( 3\times2 = 6 \) O atoms (from OH) and \( 2\times3 = 6 \) H atoms (from OH) plus \( 2\times3 = 6 \) H atoms from \( 2\text{NH}_3 \)? Wait, no, let's do H and O together. The formula for water is \( \text{H}_2\text{O} \), so H is 2 per molecule, O is 1 per molecule. On the right, \( 2\text{NH}_3 \) has \( 2\times3 = 6 \) H, \( 3\text{Mg(OH)}_2 \) has \( 3\times2 = 6 \) H (from OH) and \( 3\times2 = 6 \) O (from OH). So total H on right: \( 6 + 6 = 12 \), total O on right: 6. On the left, \( \text{Mg}_3\text{N}_2 \) has 0 H and 0 O, \( \text{H}_2\text{O} \) has 2 H and 1 O per molecule. Let x be the coefficient for \( \text{H}_2\text{O} \). Then H: \( 2x \), O: \( x \). We need \( 2x = 12 \) (for H) and \( x = 6 \) (for O). So x = 6. Let's check: \( 2x = 12 \), which matches the total H on right (12). And x = 6, so O on left is 6, which matches O on right (6). So now the equation is \( \text{Mg}_3\text{N}_2(s) + 6\text{H}_2\text{O}(l)
ightarrow 2\text{NH}_3(g) + 3\text{Mg(OH)}_2(s) \).

Let's verify all atoms:

• Mg: 3 on left, 3 on right (3×1) ✔️
• N: 2 on left, 2 on right (2×1) ✔️
• H: 6×2 = 12 on left, 2×3 (from NH3) + 3×2×2 (from Mg(OH)2? Wait no, \( 3\text{Mg(OH)}_2 \) has \( 3×2 = 6 \) H atoms (from OH) and \( 2\text{NH}_3 \) has \( 2×3 = 6 \) H atoms. So total H on right: 6 + 6 = 12. On left, \( 6\text{H}_2\text{O} \) has \( 6×2 = 12 \) H atoms. ✔️
• O: 6×1 = 6 on left (from \( 6\text{H}_2\text{O} \)), on right \( 3\text{Mg(OH)}_2 \) has \( 3×2 = 6 \) O atoms. ✔️

So the balanced equation is \( \text{Mg}_3\text{N}_2(s) + 6\text{H}_2\text{O}(l)
ightarrow 2\text{NH}_3(g) + 3\text{Mg(OH)}_2(s) \).

Answer:

The coefficients are: \( \text{Mg}_3\text{N}_2 \): 1, \( \text{H}_2\text{O} \): 6, \( \text{NH}_3 \): 2, \( \text{Mg(OH)}_2 \): 3. So filling in the "Choose..." dropdowns:

• First (Mg₃N₂): 1
• Second (H₂O): 6
• Third (NH₃): 2
• Fourth (Mg(OH)₂): 3