Question

instructions: balance the chemical equations below. identify how many atoms of each element are present once the equation is balanced.
#1
— hgo → — hg + — o2
hg: — hg: —
o: — o: —
#2
— p4 + — o2 → — p2o3
p: — p: —
o: — o: —

Explanation:

Step1: Balance equation #1

$$\text{HgO} ightarrow\text{Hg}+\text{O}_2$$

First, balance oxygen. There are 2 oxygen atoms on the right - hand side. So, put a coefficient of 2 in front of $\text{HgO}$:

$$2\text{HgO} ightarrow\text{Hg}+\text{O}_2$$

Now, balance mercury. There are 2 mercury atoms on the left - hand side. Put a coefficient of 2 in front of $\text{Hg}$:

$$2\text{HgO} ightarrow2\text{Hg}+\text{O}_2$$

Count atoms:

• Hg: 2 on left ($2\times1$ in $2\text{HgO}$) and 2 on right ($2\times1$ in $2\text{Hg}$)
• O: 2 on left ($2\times1$ in $2\text{HgO}$) and 2 on right ($1\times2$ in $\text{O}_2$)

Step2: Balance equation #2

$$\text{P}_4+\text{O}_2 ightarrow\text{P}_2\text{O}_3$$

First, balance phosphorus. There are 4 phosphorus atoms on the left. Put a coefficient of 2 in front of $\text{P}_2\text{O}_3$:

$$\text{P}_4+\text{O}_2 ightarrow2\text{P}_2\text{O}_3$$

Now, balance oxygen. There are 6 oxygen atoms on the right ($2\times3$ in $2\text{P}_2\text{O}_3$). Put a coefficient of 3 in front of $\text{O}_2$:

$$\text{P}_4 + 3\text{O}_2 ightarrow2\text{P}_2\text{O}_3$$

Count atoms:

• P: 4 on left ($1\times4$ in $\text{P}_4$) and 4 on right ($2\times2$ in $2\text{P}_2\text{O}_3$)
• O: 6 on left ($3\times2$ in $3\text{O}_2$) and 6 on right ($2\times3$ in $2\text{P}_2\text{O}_3$)

Answer:

1. $2\text{HgO}

ightarrow2\text{Hg}+\text{O}_2$ (Hg: 2, O: 2)

1. $\text{P}_4 + 3\text{O}_2

ightarrow2\text{P}_2\text{O}_3$ (P: 4, O: 6)