Final answer:
In the combustion of 10 grams of butane with 10 grams of oxygen, 0.688 moles of carbon dioxide are produced, with butane as the limiting reactant.
Step-by-step explanation:
To determine how many moles of carbon dioxide are produced from the combustion of 10 grams of butane (C₄H₁₀) with 10 grams of oxygen (O₂), it is necessary to first balance the combustion reaction:
C₄H₁₀ + 6.5O₂ → 4CO₂ + 5H₂O
Next, we convert grams to moles for each reactant.
- For butane (C₄H₁₀):
Molar mass = 58.12 g/mol
10 g C₄H₁₀ × (1 mol C₄H₁₀ / 58.12 g) = 0.172 moles of C₄H₁₀ - For oxygen (O₂):
Molar mass = 32.00 g/mol
10 g O₂ × (1 mol O₂ / 32.00 g) = 0.312 moles of O₂
The limiting reactant is the one that will run out first. Here, butane is the limiting reactant since it will produce less than 0.312 moles of oxygen needed (0.172 mol C₄H₁₀ × 6.5 mol O₂/mol C₄H₁₀ = 1.118 mol O₂ needed).
Now calculate the moles of CO₂ using the stoichiometry from the balanced equation:
0.172 moles of C₄H₁₀ × (4 moles of CO₂ / 1 mole of C₄H₁₀) = 0.688 moles of CO₂
Therefore, 0.688 moles of CO₂ are produced when 10 grams of butane burns with 10 grams of oxygen.