Question

Ethane burns in air to form carbon dioxide and water. The equation is

2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(g)

Part A

In the above equation, how many liters of oxygen (O2) are required to burn 8.0  L of ethane (C2H6) gas? Assume that both gases are measured at the same temperature and pressure.

Part B

2C2H6(g)+7O2(g)→4CO2(g)+6H2O(g)

In the equation above, what volume of ethane (C2H6) in liters is required to form 24  L of carbon dioxide (CO2)?

Part C

2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(g)

In the equation, if 14 L of ethane (C2H6) and 14 L of oxygen (O2) combined and burned to completion, which gas will be leftover after the reaction, and what is the volume of that gas remaining?

Enter the name of the gas followed by its volume in liters to the nearest whole number, separated by a comma.

Answer 1

A)

from balanced equation,

volume of O2 required = (7/2)*volume of C2H6

= (7/2)*8.0 L

= 28 L

Answer: 28 L

B)

from balanced equation,

volume of C2H6 required = (2/4)*volume of CO2

= (2/4)*24 L

= 12 L

Answer: 12 L

C)

Balanced chemical equation is:

2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(g)

2 L of C2H6 reacts with 7 L of O2

for 14 L of C2H6, 49 L of O2 is required

But we have 14 L of O2

so, O2 is limiting reagent

C2H6 will be left over

According to balanced equation

volume of C2H6 reacted = (2/7)* volume of O2

= (2/7)*14

= 4 L

volume of C2H6 remaining = volume initially present - volume reacted

= 14 - 4

= 10 L

Answer: ethane, 10