To determine the mole ratio of [tex]\(\text{PbO}_2\)[/tex] (lead dioxide) to [tex]\(\text{H}_2\text{O}\)[/tex] (water) in the given chemical reaction, follow these steps:
1. Identify the given balanced chemical equation:
[tex]\[
\text{Pb} + \text{PbO}_2 + 2\text{H}_2\text{SO}_4 \rightarrow 2\text{PbSO}_4 + 2\text{H}_2\text{O}
\][/tex]
2. Determine the coefficients of the reactants and products from the balanced equation:
- Notice the coefficient for [tex]\(\text{PbO}_2\)[/tex] is [tex]\(1\)[/tex].
- Notice the coefficient for [tex]\(\text{H}_2\text{O}\)[/tex] is [tex]\(2\)[/tex].
3. Establish the mole ratio based on the coefficients:
- For every 1 mole of [tex]\(\text{PbO}_2\)[/tex] consumed, 2 moles of [tex]\(\text{H}_2\text{O}\)[/tex] are produced.
- This can be represented as:
[tex]\[
\text{PbO}_2 : \text{H}_2\text{O} = 1 : 2
\][/tex]
4. Write the mole ratio explicitly:
[tex]\[
\text{PbO}_2 : \text{H}_2\text{O} = 1 : 2
\][/tex]
Therefore, the mole ratio of [tex]\(\text{PbO}_2\)[/tex] to [tex]\(\text{H}_2\text{O}\)[/tex] is [tex]\(1 : 2\)[/tex].
So, the final answer is:
[tex]\[
PbO_2 : H_2O = 1 : 2
\][/tex]
