Final answer:
To determine the grams of water formed when 93.72 g of propane (C3H8) is burned in excess oxygen, we need to use the balanced equation and stoichiometry. The answer is approximately 114.9 grams of water formed (c).
Step-by-step explanation:
To determine the grams of water formed when 93.72 g of propane (C3H8) is burned in excess oxygen, we need to use the balanced equation and stoichiometry. From the balanced equation, we can see that for every mole of propane, 4 moles of water are formed. To calculate the moles of propane, we divide the mass of propane by the molar mass. Then, we multiply the moles of propane by the mole ratio to find the moles of water formed. Finally, we convert the moles of water to grams by multiplying by the molar mass of water.
Step-by-step calculation:
1. Moles of propane = mass of propane / molar mass of propane
2. Moles of water formed = moles of propane * mole ratio (4 moles of water per 1 mole of propane)
3. Grams of water formed = moles of water formed * molar mass of water
Using this method, we find that the answer is approximately 114.9 grams of water formed.