When 20.0 g of C2H6 and 60.0 g of O2 react completely to form CO2 and H2O, what is the limiting reactant?

• A. O2
• B. CO2
• C. C2H6
• D. H2O

Answer: (C)

Verified step by step guidance

1. Write the balanced chemical equation for the combustion of ethane (C2H6): \(\mathrm{2\ C_2H_6 + 7\ O_2 \rightarrow 4\ CO_2 + 6\ H_2O}\).
2. Calculate the number of moles of each reactant using their given masses and molar masses: \(\text{moles of } C_2H_6 = \frac{20.0\ \mathrm{g}}{30.07\ \mathrm{g/mol}}\) and \(\text{moles of } O_2 = \frac{60.0\ \mathrm{g}}{32.00\ \mathrm{g/mol}}\).
3. Use the mole ratio from the balanced equation to find the amount of \(O_2\) required to completely react with the given moles of \(C_2H_6\): \(\text{moles of } O_2 \text{ required} = \text{moles of } C_2H_6 \times \frac{7}{2}\).
4. Compare the actual moles of \(O_2\) available with the moles of \(O_2\) required to determine if \(O_2\) is in excess or limiting.
5. Identify the limiting reactant as the one that produces the least amount of product or is completely consumed first; in this case, if \(O_2\) is in excess, then \(C_2H_6\) is the limiting reactant.