Calculate the total volume of residual gases after the reaction between 60 cm³ of H₂ and 20 cm³ of O₂ at 100°C.

The reaction of hydrogen gas with oxygen gas produces water. Since the temperature is $100^\circ\text{C}$, water will be in its gaseous state (steam).

According to Gay-Lussac's Law of Combining Volumes, at constant temperature and pressure, the volumes of gaseous reactants and products are in simple whole-number ratios corresponding to the stoichiometric coefficients in the balanced equation.
From the equation, 2 volumes of $\text{H}_2$ react with 1 volume of $\text{O}_2$.

Given volumes:
Volume of $\text{H}_2 = 60 \text{ cm}^3$
Volume of $\text{O}_2 = 20 \text{ cm}^3$

To react completely with $20 \text{ cm}^3$ of $\text{O}_2$, the required volume of $\text{H}_2$ would be:

Since we have $60 \text{ cm}^3$ of $\text{H}_2$, which is more than the $40 \text{ cm}^3$ required, oxygen ($\text{O}_2$) is the limiting reactant.

• Oxygen consumed: All $20 \text{ cm}^3$ of $\text{O}_2$ will be consumed.
• Hydrogen consumed: Based on the $2:1$ ratio with $\text{O}_2$, $2 \times 20 \text{ cm}^3 = 40 \text{ cm}^3$ of $\text{H}_2$ will be consumed.
• Water vapor formed: Based on the $2:2$ (or $1:1$) ratio with $\text{O}_2$, $2 \times 20 \text{ cm}^3 = 40 \text{ cm}^3$ of $\text{H}_2\text{O}(\text{g})$ will be formed.

• Unreacted hydrogen: Initial $\text{H}_2$ - Consumed $\text{H}_2 = 60 \text{ cm}^3 - 40 \text{ cm}^3 = 20 \text{ cm}^3$.
• Unreacted oxygen: $0 \text{ cm}^3$ (all consumed).
• Water vapor (product): $40 \text{ cm}^3$.

The residual gases are the unreacted hydrogen and the water vapor formed.

Total volume of residual gases = Volume of unreacted $\text{H}_2$ + Volume of $\text{H}_2\text{O}(\text{g})$
Total volume of residual gases = $20 \text{ cm}^3 + 40 \text{ cm}^3 = 60 \text{ cm}^3$.

The total volume of the residual gases is $60 \text{ cm}^3$.

The final answer is $\text{A. } 60 \text{ cm}^3$.