When 64.0 g of C_2H_2 (acetylene) are completely combusted in oxygen, how many moles of H_2O are produced?

• A. 3.56 moles
• B. 3.20 moles
• C. 3.56 moles
• D. 3.20 moles

Answer: (D)

Verified step by step guidance

1. Write the balanced chemical equation for the complete combustion of acetylene (C\_2H\_2). The general form for combustion of a hydrocarbon is: \(\text{C}_x\text{H}_y + O_2 \rightarrow CO_2 + H_2O\) Balance the equation for C\_2H\_2 combustion.
2. Calculate the molar mass of acetylene (C\_2H\_2) by summing the atomic masses: \(\text{Molar mass} = 2 \times \text{atomic mass of C} + 2 \times \text{atomic mass of H}\).
3. Convert the given mass of acetylene (64.0 g) to moles using the molar mass: \(\text{moles of C}_2\text{H}_2 = \frac{64.0 \text{ g}}{\text{molar mass of C}_2\text{H}_2}\).
4. Use the mole ratio from the balanced equation between C\_2H\_2 and H\_2O to find the moles of water produced: \(\text{moles of H}_2\text{O} = \text{moles of C}_2\text{H}_2 \times \frac{\text{moles of H}_2\text{O}}{\text{moles of C}_2\text{H}_2}\) (from the balanced equation).
5. Report the calculated moles of H\_2O as the final answer, which corresponds to the amount of water produced from the complete combustion of 64.0 g of acetylene.