Answer:
- To determine which reactant is limiting, we need to compare the moles of each reactant and the mole ratio of oxygen to pentane in the balanced chemical equation:
C5H12 + 8 O2 → 5 CO2 + 6 H2O
Moles of pentane = 35.45 g / 72.15 g/mol = 0.491 mol
Moles of oxygen = 25 g / 32 g/mol = 0.781 mol
Mole ratio of oxygen to pentane is 8:1, so we need 8 moles of oxygen for every mole of pentane. Since we have 0.491 moles of pentane, we need 3.928 moles of oxygen. However, we only have 0.781 moles of oxygen, which means oxygen is the limiting reactant.
To calculate the grams of oxygen that react, we can use the mole ratio of oxygen to pentane:
0.491 mol C5H12 × (8 mol O2 / 1 mol C5H12) × (32 g/mol) = 125.44 g O2
The amount of excess reactant left over can be calculated by subtracting the moles of the limiting reactant from the moles of the excess reactant:
Moles of excess oxygen = 0.781 mol O2 - 0.0614 mol O2 = 0.7196 mol O2
Grams of excess oxygen = 0.7196 mol × 32 g/mol = 23.03 g O2
Therefore, 125.44 grams of oxygen react, and 23.03 grams of oxygen are left over.
2. The amount of energy released can be calculated by multiplying the amount of moles of pentane by the heat of combustion:
0.491 mol C5H12 × (-173.5 kJ/mol) = -85.17 kJ
Therefore, 85.17 kJ of energy is released during the combustion reaction.