Given the balanced equation for the combustion of octane: 2 C_8H_{18} + 25 O_2 → 16 CO_2 + 18 H_2O, if 100.0 g of C_8H_{18} is mixed with 400.0 g of O_2 and allowed to react completely, how much octane (C_8H_{18}) remains unreacted after the reaction?

• A. 25.0 g
• B. 50.0 g
• C. 12.5 g
• D. 0.0 g

Answer: (D)

Verified step by step guidance

1. Identify the balanced chemical equation: \$2\ C_8H_{18} + 25\ O_2 \rightarrow 16\ CO_2 + 18\ H_2O$.
2. Calculate the molar masses of octane (\(C_8H_{18}\)) and oxygen (\(O_2\)). For octane, use \$8 \times 12.01 + 18 \times 1.008\(, and for oxygen, use \)2 \times 16.00$ g/mol.
3. Convert the given masses of octane (100.0 g) and oxygen (400.0 g) to moles by dividing each mass by its respective molar mass.
4. Determine the limiting reactant by comparing the mole ratio of the reactants to the stoichiometric ratio from the balanced equation. Calculate how many moles of oxygen are required to react with the available moles of octane, and vice versa.
5. Since the limiting reactant is the one that will be completely consumed, calculate the amount of octane that reacts based on the limiting reactant. Subtract this reacted amount from the initial 100.0 g of octane to find the remaining unreacted octane.