Question

Gaseous ethane (CH,CH,) will react with gaseous oxygen (0,) to produce gaseous carbon dioxide (CO) and gaseous water (H2O). Suppose 1.50 g of

ethane is mixed with 11. g of oxygen. Calculate the minimum mass of ethane that could be left over by the chemical reaction. Round your answer to 2
significant digits.
​

Answer 1

0.00 g

Step-by-step explanation:

We have the masses of two reactants, so this is a limiting reactant problem.

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.

Mᵣ 30.07 32.00

2CH₃CH₃ + 7O₂ ⟶ 4CO₂ + 6H₂O

Mass/g: 1.50 11.

2. Calculate the moles of each reactant

`\text{moles of C$_(2)$H}_(6) = \text{1.50 g C$_(2)$H}_(6) * \frac{\text{1 mol C$_(2)$H}_(6)}{\text{30.07 g C$_(2)$H}_(6)} = \text{0.04988 mol C$_(2)$H}_(6)\\\\\text{moles of O}_(2) = \text{11. g O}_(2) * \frac{\text{1 mol O}_(2)}{\text{32.00 g O}_(2)} = \text{0.34 mol O}_(2)`

3. Calculate the moles of CO₂ we can obtain from each reactant

From ethane:

The molar ratio is 4 mol CO₂:2 mol C₂H₆

`\text{Moles of CO}_(2) = \text{0.04988 mol C$_(2)$H}_(6) * \frac{\text{4 mol CO}_(2)}{\text{2 mol C$_(2)$H}_(6)} = \text{0.09976 mol CO}_(2)`

From oxygen:

The molar ratio is 4 mol CO₂:7 mol O₂

`\text{Moles of CO}_(2) = \text{0.34 mol O}_(2)* \frac{\text{4 mol CO}_(2)}{\text{7 mol O}_(2)} = \text{0.20 mol CO}_(2)`

4. Identify the limiting and excess reactants

The limiting reactant is ethane, because it gives the smaller amount of CO₂.

The excess reactant is oxygen.

5. Mass of ethane left over.

Ethane is the limiting reactant. It will be completely used up.

The mass of ethane left over will be 0.00 g.