Final answer:
The complete combustion of 20 cm³ of butane in excess oxygen produces 180 cm³ of carbon dioxide and water vapor. Considering the unreacted oxygen, the total volume of gas after the reaction is 200 cm³. Hence, the correct answer is 200 cm³, corresponding to option D.
Step-by-step explanation:
The question concerns the complete combustion of butane (C4H10) in the presence of oxygen (O2) to form carbon dioxide (CO2) and water vapor (H2O). The provided reaction is incorrect; the correct balanced equation should be:
2 C4H10(g) + 13 O2(g) → 8 CO2(g) + 10 H2O(g)
According to the stoichiometry of the reaction, the combustion of butane requires 13 moles of oxygen to produce 8 moles of carbon dioxide and 10 moles of water.
Given that 20 cm³ of butane is burned in 150 cm³ of oxygen, we will first balance the volumes on a molar basis:
- 2 volumes of butane react with 13 volumes of oxygen.
- Since we have 20 cm³ of butane, the proportional volume of oxygen required would be 20 cm³ × (13/2), which equals 130 cm³.
- However, we only have 150 cm³ of oxygen, which is sufficient to completely combust the butane.
- The combustion will produce 8 volumes of CO2 and 10 volumes of water vapor for every 2 volumes of butane; which results in (20 cm³ × (8/2)) + (20 cm³ × (10/2)) = 80 cm³ + 100 cm³ = 180 cm³ of products.
Since the oxygen was in excess, the remaining unreacted oxygen volume is 150 cm³ - 130 cm³ = 20 cm³. Thus, the total volume of gases after combustion is 180 cm³ of products plus 20 cm³ of unreacted oxygen, equaling 200 cm³.
Therefore, the answer is D) 200 cm³.