When 44 grams of propane (C_3H_8) are completely combusted in excess oxygen, how many grams of water (H_2O) are produced?

• A. 18 grams
• B. 88 grams
• C. 54 grams
• D. 36 grams

Answer: (C)

Verified step by step guidance

1. Write the balanced chemical equation for the combustion of propane (C\_3H\_8): \(\mathrm{C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O}\)
2. Calculate the molar mass of propane (C\_3H\_8) by summing the atomic masses: Carbon (C) = 12 g/mol, Hydrogen (H) = 1 g/mol, so \(\mathrm{Molar\ mass\ of\ C_3H_8 = (3 \times 12) + (8 \times 1) = 44\ g/mol}\)
3. Determine the number of moles of propane combusted using the given mass: \(\mathrm{moles\ of\ C_3H_8 = \frac{44\ g}{44\ g/mol} = 1\ mol}\)
4. Use the stoichiometric ratio from the balanced equation to find moles of water produced: For every 1 mole of C\_3H\_8, 4 moles of H\_2O are produced, so \(\mathrm{moles\ of\ H_2O = 1 \times 4 = 4\ moles}\)
5. Calculate the mass of water produced by multiplying moles of water by its molar mass (H\_2O = 18 g/mol): \(\mathrm{mass\ of\ H_2O = 4\ moles \times 18\ g/mol}\)