Question

1. _ (nh₄)₃po₄ + _ pb(no₃)₄ → _ pb₃(po₄)₄ + _ nh₄no₃
2. _ secl₆ + _ o₂ → _ seo₂ + _ cl₂

Explanation:

Step1: Balance problem 9 (Pb atoms)

Count Pb: reactant has 1, product has 3. Multiply $\text{Pb(NO}_3\text{)}_4$ by 3.

$$3\text{Pb(NO}_3\text{)}_4 + \text{(NH}_4\text{)}_3\text{PO}_4 ightarrow \text{Pb}_3\text{(PO}_4\text{)}_4 + \text{NH}_4\text{NO}_3$$

Step2: Balance problem 9 (PO₄ groups)

Count $\text{PO}_4$: reactant has 1, product has 4. Multiply $\text{(NH}_4\text{)}_3\text{PO}_4$ by 4.

$$3\text{Pb(NO}_3\text{)}_4 + 4\text{(NH}_4\text{)}_3\text{PO}_4 ightarrow \text{Pb}_3\text{(PO}_4\text{)}_4 + \text{NH}_4\text{NO}_3$$

Step3: Balance problem 9 (NH₄⁺ groups)

Count $\text{NH}_4^+$: reactant has 12, product has 1. Multiply $\text{NH}_4\text{NO}_3$ by 12.

$$3\text{Pb(NO}_3\text{)}_4 + 4\text{(NH}_4\text{)}_3\text{PO}_4 ightarrow \text{Pb}_3\text{(PO}_4\text{)}_4 + 12\text{NH}_4\text{NO}_3$$

Step4: Verify problem 9 (NO₃⁻ groups)

Count $\text{NO}_3^-$: reactant has 12, product has 12. Equation is balanced.

Step5: Balance problem 10 (Se atoms)

Se is balanced (1 on each side).

$$\text{SeCl}_6 + \text{O}_2 ightarrow \text{SeO}_2 + \text{Cl}_2$$

Step6: Balance problem 10 (Cl atoms)

Count Cl: reactant has 6, product has 2. Multiply $\text{Cl}_2$ by 3.

$$\text{SeCl}_6 + \text{O}_2 ightarrow \text{SeO}_2 + 3\text{Cl}_2$$

Step7: Verify problem 10 (O atoms)

Count O: reactant has 2, product has 2. Equation is balanced.

Answer:

1. $4\text{(NH}_4\text{)}_3\text{PO}_4 + 3\text{Pb(NO}_3\text{)}_4

ightarrow \text{Pb}_3\text{(PO}_4\text{)}_4 + 12\text{NH}_4\text{NO}_3$

1. $\text{SeCl}_6 + \text{O}_2

ightarrow \text{SeO}_2 + 3\text{Cl}_2$