Question

1. _ cs + _ n₂ → ___ cs₃n
2. _ mg + _ cl₂ → ___ mgcl₂
3. _ rb + _ rbno₃ → _ rb₂o + _ n₂
4. _ h₃po₄ + _ ca(oh)₂ → _ ca₃(po₄)₂ + _ h₂o
5. _ n₂ + _ h₂ → ___ nh₃

Explanation:

Step1: Balance equation 14

For \( \text{Cs}+\text{N}_2
ightarrow\text{Cs}_3\text{N} \), balance \( N \) first. There are \( 2 \) \( N \) atoms on the left and \( 1 \) on the right. Multiply \( \text{Cs}_3\text{N} \) by \( 2 \). Then balance \( Cs \): \( 6\text{Cs}+\text{N}_2 = 2\text{Cs}_3\text{N} \)

Step2: Balance equation 15

For \( \text{Mg}+\text{Cl}_2
ightarrow\text{MgCl}_2 \), count atoms. \( 1 \) \( Mg \) and \( 2 \) \( Cl \) on both sides. So \( \text{Mg}+\text{Cl}_2=\text{MgCl}_2 \)

Step3: Balance equation 16

For \( \text{Rb}+\text{RbNO}_3
ightarrow\text{Rb}_2\text{O}+\text{N}_2 \), balance \( N \): multiply \( \text{RbNO}_3 \) by \( 2 \), then \( \text{N}_2 \) by \( 1 \). Balance \( O \): \( 2\text{RbNO}_3 \) gives \( 6 \) \( O \), so \( 6\text{Rb}_2\text{O} \). Then balance \( Rb \): \( 10\text{Rb}+2\text{RbNO}_3 = 6\text{Rb}_2\text{O}+\text{N}_2 \)

Step4: Balance equation 17

For \( \text{H}_3\text{PO}_4+\text{Ca(OH)}_2
ightarrow\text{Ca}_3(\text{PO}_4)_2+\text{H}_2\text{O} \), balance \( Ca \): multiply \( \text{Ca(OH)}_2 \) by \( 3 \). Balance \( P \): multiply \( \text{H}_3\text{PO}_4 \) by \( 2 \). Then balance \( H \) and \( O \): \( 2\text{H}_3\text{PO}_4 + 3\text{Ca(OH)}_2=\text{Ca}_3(\text{PO}_4)_2+6\text{H}_2\text{O} \)

Step5: Balance equation 18

For \( \text{N}_2+\text{H}_2
ightarrow\text{NH}_3 \), balance \( N \): multiply \( \text{NH}_3 \) by \( 2 \). Then balance \( H \): \( \text{N}_2 + 3\text{H}_2=2\text{NH}_3 \)

Answer:

1. \( 6\text{Cs}+\text{N}_2 = 2\text{Cs}_3\text{N} \)
2. \( \text{Mg}+\text{Cl}_2=\text{MgCl}_2 \)
3. \( 10\text{Rb}+2\text{RbNO}_3 = 6\text{Rb}_2\text{O}+\text{N}_2 \)
4. \( 2\text{H}_3\text{PO}_4 + 3\text{Ca(OH)}_2=\text{Ca}_3(\text{PO}_4)_2+6\text{H}_2\text{O} \)
5. \( \text{N}_2 + 3\text{H}_2=2\text{NH}_3 \)