When 20.0 g of propane (C_3H_8) is completely combusted in excess oxygen, how many grams of water (H_2O) are produced?

• A. 18.0 g
• B. 32.7 g
• C. 20.0 g
• D. 44.0 g

Answer: (B)

Verified step by step guidance

1. Write the balanced chemical equation for the complete combustion of propane (C\_3H\_8): \(\mathrm{C\_3H\_8 + 5 O\_2 \rightarrow 3 CO\_2 + 4 H\_2O}\)
2. Calculate the molar mass of propane (C\_3H\_8) by summing the atomic masses: 3 carbons and 8 hydrogens.
3. Convert the given mass of propane (20.0 g) to moles using the molar mass calculated: \(moles\ of\ C\_3H\_8 = \frac{20.0\ g}{molar\ mass\ of\ C\_3H\_8}\)
4. Use the mole ratio from the balanced equation to find moles of water produced. For every 1 mole of propane combusted, 4 moles of water are formed: \(moles\ of\ H\_2O = moles\ of\ C\_3H\_8 \times 4\)
5. Convert the moles of water to grams by multiplying by the molar mass of water (H\_2O): \(mass\ of\ H\_2O = moles\ of\ H\_2O \times 18.0\ g/mol\)