Gaseous ethane CH3CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. Suppose 2.1 g of ethane is mixed with 13.2 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.

Question

Dayton

Chemistry

27 March, 01:09

Gaseous ethane CH3CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. Suppose 2.1 g of ethane is mixed with 13.2 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.

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Answers (2)

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Taryn Mora · Answer 1

Since ethane is the limiting reactant, it will completely be consumed. 0 grams will left over. O2 is in excess, there will remain 5.4 grams of oxygen

Explanation:

Step 1: Data given

Mass of ethane = 2.1 grams

Mass of oxygen = 13.2 grams

Molar mass of ethane = 30.07 g/mol

Molar mass of oxygen = 32.0 g/mol

Step 2: The balanced equation

2C2H6 + 7O2 → 4CO2 + 6H2O

Step 3: Calculate moles ethane

Moles ethane = 2.1 grams / 30.07 g/mol

Moles ethane = 0.0698 moles

Step 4: Calculate moles oxygen

Moles O2 = 13.2 grams / 32.0 g/mol

Moles O2 = 0.4125 moles

Step 5: Calculate the limiting reactant

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

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

There will remain 0.4125 - 0.2443 = 0.1682 moles

Step 6: calculate the mass oxygen remaining

Mass O2 = 0.1682 moles * 32.0 g/mol

Mass O2 = 5.4 grams

Since ethane is the limiting reactant, it will completely be consumed. 0 grams will left over. O2 is in excess, there will remain 5.4 grams of oxygen

Cantu · Answer 2

All the ethane will be consumed by the chemical reaction, so no grams to react will be left over. This occurs because the ethanol is the limiting reagent.

Explanation:

Reactions that occur with a compound and oxygen are called combustion, and the products are always CO₂ and water.

The equation for the combustion of ethane is:

2CH₃CH₃ (g) + 7O₂ (g) → 4CO₂ (g) + 6H₂O (g)

We convert the mass of the reactants, in order to find out the moles of each.

2.1 g. 1 mol / 30 g = 0.07 moles

13.2 g / 32 g/mol = 0.412 moles

We now propose the following rule of three:

7 moles of oxgen need 2 moles of ethane to react

Then, 0.412 moles of O₂ will react with (0.412. 2) / 7 = 0.118 moles of ethane. We only have 0.07 moles, so we don't have enough ethane. That's why it is the limiting reactant.

All the ethane will be consumed by the chemical reaction, so no grams will be left over.