Explanation:
2 C₆H₁₀ + 17 O₂ ---> 12 CO₂ + 10 H₂O
With start with 35.0 g of C₆H₁₀ and 40.0 g of O₂. The first thing that we have to do is to convert those grams into moles using their molar masses.
atomic mass of C = 12.01 amu
atomic mass of H = 1.01 amu
atomic mass of O = 16.00 amu
molar mass of C₆H₁₀ = 6 * 12.01 g/mol + 10 * 1.01 g/mol
molar mass of C₆H₁₀ = 82.16 g/mol
molar mass of O₂ = 2 * 16.00 g/mol
molar mass of O₂ = 32.00 g/mol
mass of C₆H₁₀ = 35.0 g
moles of C₆H₁₀ = 35.0 g/(82.16 g/mol)
moles of C₆H₁₀ = 0.426 moles
mass of O₂ = 40.0 g
moles of O₂ = 40.0 g/(32.00 g/mol)
moles of O₂ = 1.25 moles
2 C₆H₁₀ + 17 O₂ ---> 12 CO₂ + 10 H₂O
According to the coefficients of the reaction 2 moles of C₆H₁₀ will react with 17 moles of O₂. The molar ratio between them is 2 to 17.
2 moles of C₆H₁₀ = 17 moles of O₂
We mixed 0.426 moles of C₆H₁₀ with only 1.25 moles of O₂. Since the relationship between them is 2 to 17, it seems that the O₂ is the limiting reactant. But let's check it, let's find the number of moles of C₆H₁₀ that will completely react with 1.25 moles of O₂.
moles of C₆H₁₀ = 1.25 moles of O₂ * 2 moles of C₆H₁₀/(17 moles of O₂)
moles of C₆H₁₀ = 0.147 moles
We found that 1.25 moles of O₂ will completely react with only 0.147 moles of C₆H₁₀. We mixed 1.25 moles of O₂ and 0.426 moles of C₆H₁₀. So C₆H₁₀ is in excess and O₂ is the limiting reactant.
limiting reactant = O₂
excess reactant = C₆H₁₀
Once we found that the limiting reactant is O₂ we can find the number of moles of CO₂ produced by 1.25 moles of O₂. And finally convert those moles of CO₂ back to grams to find the answer to our problem.
2 C₆H₁₀ + 17 O₂ ---> 12 CO₂ + 10 H₂O
17 moles of O₂ = 12 moles of CO₂
moles of CO₂ produced = 1.25 moles of O₂ * 12 moles of CO₂/(17 moles of O₂)
moles of CO₂ produced = 0.882 moles
atomic mass of C = 12.01 amu
atomic mass of O = 16.00 amu
molar mass of CO₂ = 12.01 * 1 + 16.00 * 2
molar mass of CO₂ = 44.01 g/mol
mass of CO₂ produced = 0.882 moles * 44.01 g/mol
mass of CO₂ produced = 38.8 g
Answer:
grams of CO₂ formed = 38.8 g
limiting reactant = O₂
excess reactant = C₆H₁₀