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The following reaction occurs:2 C6H10 + 17 O2 ---> 12 CO2 + 10 H2OIf I start with 35.0 grams of C6H10 and 40.0 grams of O2. Determine how many grams of CO2 will be formed and what is the limiting reactant and excess reactant

The Following Reaction Occurs2 C6H10 17 O2 Gt 12 CO2 10 H2OIf I Start With 350 Grams Of C6H10 And 400 Grams Of O2 Determine How Many Grams Of CO2 Will Be Formed class=

Sagot :

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

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