Question

What are \(x\) and \(y\) in the following reaction?
\[ x\mathrm{Pb_3O_4}\rightarrow y\mathrm{PbO}+O_2 \]

• A. \(x=3,\ y=6\)
• B. \(x=2,\ y=4\)
• C. \(x=2,\ y=5\)
• D. \(x=2,\ y=6\)

Hint:

Always balance metal atoms first and then balance oxygen or hydrogen atoms at the end.

Answer 1

The Correct Option is D

Step 1: Write the skeleton equation.
The decomposition to balance is \[ x\,Pb_3O_4 \rightarrow y\,PbO + O_2 \] and we must find whole number values of $x$ and $y$.
Step 2: Count atoms on each side per formula unit.
One $Pb_3O_4$ has $3$ lead and $4$ oxygen atoms, while one $PbO$ has $1$ lead and $1$ oxygen, and $O_2$ has $2$ oxygen atoms.
Step 3: Balance lead first.
Choosing $x = 2$ gives $2 \times 3 = 6$ lead atoms on the left, so the right side needs $6$ lead atoms, which means $y = 6$.
Step 4: Check oxygen.
Left oxygen is $2 \times 4 = 8$. Right oxygen is $6$ from $6\,PbO$ plus $2$ from $O_2$, giving $6 + 2 = 8$. The oxygen balances perfectly.
Step 5: Write the balanced equation.
\[ 2\,Pb_3O_4 \rightarrow 6\,PbO + O_2 \]
Step 6: State the values.
Thus $x = 2$ and $y = 6$, which is option 4.
\[ \boxed{x=2,\ y=6} \]