The combustion of glucose yields carbon dioxide and water according to the following reaction: C₆H₁₂O₆ (s) + 6 O₂ (g) → 6 CO₂ (g) + 6 H₂O (l) Calculate the mass of H₂O produced when 12.0 g of glucose is combusted in a container of oxygen with a volume of 1.3 L at a pressure of 2.2 atm and a temperature of 205℃.

• A. 0.20 g
• B. 1.2 g
• C. 1.3 g
• D. 7.2 g
• E. 7.9 g

Answer: (C)

Verified step by step guidance

1. Step 1: Begin by identifying the molar mass of glucose (C6H12O6). Calculate the molar mass using the atomic masses of carbon (C), hydrogen (H), and oxygen (O). The molar mass of glucose is calculated as follows: \( \text{Molar mass of } C_6H_{12}O_6 = 6 \times \text{atomic mass of C} + 12 \times \text{atomic mass of H} + 6 \times \text{atomic mass of O} \).
2. Step 2: Convert the mass of glucose given (12.0 g) into moles using its molar mass. Use the formula: \( \text{moles of glucose} = \frac{\text{mass of glucose}}{\text{molar mass of glucose}} \).
3. Step 3: Use the stoichiometry of the balanced chemical equation to determine the moles of water (H2O) produced. According to the equation, 1 mole of glucose produces 6 moles of water. Therefore, multiply the moles of glucose by 6 to find the moles of water produced.
4. Step 4: Calculate the mass of water produced using the moles of water and the molar mass of water (H2O). The molar mass of water is calculated as: \( \text{Molar mass of } H_2O = 2 \times \text{atomic mass of H} + \text{atomic mass of O} \). Use the formula: \( \text{mass of water} = \text{moles of water} \times \text{molar mass of water} \).
5. Step 5: Consider the conditions of the container with oxygen (volume, pressure, and temperature) to ensure there is sufficient oxygen for complete combustion. Use the ideal gas law \( PV = nRT \) to calculate the moles of oxygen available and verify that it exceeds the stoichiometric requirement for the reaction.