Pentane, C5H12, will react as shown below:
____C5H12(l) + ____O2 → ______CO2 + ______H2O
1. Balance the equation above by putting coefficients into each blank space.
- Answer: 1C5H12 + 8O2 → 5CO2 + 6H2O
2. How many atoms of oxygen are in 144 grams of pentane?
Step 1: 144 g of C5H12 × (1 mole / 72.17 g) = 1.995 moles of C5H12
Step 2: 1.995 moles of C5H12 × (8 moles of O2 / 1 mole of C5H12) = 15.962 moles of O2
Step 3: 15.962 moles of O2 × (6.02 × 10^23 atoms of O2 / 1 mole of O2) = 9.609 × 10^24 atoms of O2
3. How many atoms of oxygen are in 192 grams of carbon dioxide gas?
Step 1: 192 grams of CO2 × (1 mole of CO2 / 44.01 grams of CO2) = 4.363 moles of CO2
Step 2: 4.363 moles of CO2 × (2 atoms of O / 1 mole of CO2) = 8.726 moles of O atoms
Step 3: 8.726 moles of O atoms × (6.02 × 10^23 atoms / 1 mole) = 5.253 × 10^24 atoms of O
4. A) CIRCLE ONE: True or False:
The reactant you begin with fewer grams of is your limiting reactant.
B) If 32 g of oxygen gas and 32 g of pentane are combined, how much water will be produced, in moles? (Use your reaction from the top of the first page)
Note: Review the calculations for accuracy and ensure the steps align with stoichiometric principles.
Answer (2)
Final Answers:
Balanced Chemical Equation: 1 C₅ H 12 (I) + 8 O₂ → 5 CO₂+ 6 H₂O(g)
Atoms of Oxygen in 144 grams of Pentane: 9.609 × 1 0 24 atoms of O 2
Atoms of Oxygen in 192 grams of CO2: 2.627 × 1 0 24 atoms of O 2
A) True
B) Water produced: 2.5 moles ;
The balanced equation for the combustion of pentane is: 1 C₅H₁₂ + 8 O₂ → 5 CO₂ + 6 H₂O. In 144 grams of pentane, there are approximately 9.609 × 10²⁴ atoms of O, and from 192 grams of CO₂, there are roughly 5.253 × 10²⁴ atoms of O. The limiting reactant equation exercise reveals that 32 grams of C₅H₁₂ yields about 2.66 moles of water.
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