Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water. Suppose 2.10 g of ethane is mixed with 12. g of oxygen. Calculate the minimum mass of ethane that could be left over by the chemical reaction. Be sure your answer has the correct number of significant digits.

Question

Chemistry

Kaylee Velazquez

13 May, 13:10

Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water. Suppose 2.10 g of ethane is mixed with 12. g of oxygen. Calculate the minimum mass of ethane that could be left over by the chemical reaction. Be sure your answer has the correct number of significant digits.

+5

Answers (2)

Know the Answer?

Draven Hopkins · Answer 1

Since ethane is the limting reactant, there will remain nothing. The mass ethane remaingin is 0 grams

There remains 4.18 grams of oxygen

Explanation:

Step 1: data given

Ethane = C2H6

Mass of ethane = 2.10 grams

Molar mass C2H6 = 30.07 g/mol

oxygen = O2

Mass of oxygen = 12.0 grams

Molar mass O2 = 32.0 g/mol

Step 2: The balanced equation

2C2H6 + 7O2 → 4CO2 + 6H2O

Step 3: Calculate moles C2H6

Moles C2H6 = mass C2H6 / molar mass C2H6

Moles C2H6 = 2.10 grams / 30.07 g/mol

Moles C2H6 = 0.0698 moles

Step 4: Calculate moles O2

Moles O2 = 12.0 grams / 32.0 g/mol

Moles O2 = 0.375 moles

Step 5: Calculate limiting reactant

For 2 moles ethane we have 7 moles O2 to produce 4 moles CO2 and 6 moles H2O

Ethane is the limiting reactant. It wil completely be consumed (0.0698 moles). O2 is in excess. There will react 3.5 * 0.0698 = 0.2443 moles

There will remain 0.375 - 0.2443 = 0.1307 moles

Since ethane is the limting reactant, there will remain nothing. The mass ethane remaingin is 0 grams

Step 6: Calculate mass oxygen remaining

Mass oxygen = 0.1307 moles * 32.0 g/mol

Mass oxygen remaining = 4.18 grams

Alannah · Answer 2

Ethane is the limiting reactant, so no mass of ethane will be left over by the chemical reaction. All the mass will react, the 2.1 grams of ethane.

Explanation:

First of all, we need to determine the reaction and the limiting reactant to work with the stoichiometry.

The equation is: 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O

We define the moles of the reactants:

2.10 g / 30 g/mol = 0.07 moles of ethane

12 g / 32 g/mol = 0.375 moles of oxygen

To determine the limiting reactant, we start with oxygen:

7 moles of O₂ can react with 2 moles of ethane

Then, 0.375 moles of O₂ will react with (0.375. 2) / 7 = 1.31 moles of ethane.

We do not have enough ethane, just only 0.07 moles to react.

Ethane is the limiting reactant, so no mass of ethane will be left over by the chemical reaction. All the mass will react, the 2.1 grams of ethane.