When ethane (C2H6) burns, it produces carbon dioxide and water:

2C2H6 (g) + 7O2 (g) →4CO2 (g) + 6H2O (l)

How many liters of carbon dioxide will be produced when 89.5 L of ethane are burned?

(One mole of any gas occupies 22.4 L under certain conditions of temperature and pressure. Assume those conditions for this question.)

179 L

313 L

358 L

538 L

179 L

Explanation:

We are given;

The equation for the combustion of ethane.;

2C₂H₆ (g) + 7O₂g) →4CO₂ (g) + 6H₂O (l)

Volume of ethane gas used as 89.5 L

We are required to determine the volume of carbon dioxide produced;

Step 1: Determine the number of moles of ethane gas

According to the molar gas volume, 1 mole of a gas occupies a volume of 22.4 L at STP.

Therefore;

Number of moles of ethane = Volume given/molar gas volume

Thus;

Moles of ethane = 89.5 L : 22.4 L

= 3.9955

= 4.0 moles

Step 2: Determine the number of moles of CO₂ produced

From the equation,

12 moles of ethane reacts to produce 4 moles of carbon dioxide

Thus; Moles of CO₂ = Moles of ethane * 2

= 4.0 moles * 2

= 8.0 moles

Step 3: Volume of CO₂ produced

We know that;

1 mole of a gas = 22.4 L at STP

Therefore;

Volume of CO₂ = 8.0 moles * 22.4 L/mol

= 179.2 L

= 179 L

Therefore, the volume of CO₂ is 179 L