If 5.49 mol of ethane (#C_2H_6#) undergoes combustion according to the unbalanced equation #C_2H_6+O_2-CO_2+H_2O#, how much oxygen is required?

Question

If 5.49 mol of ethane (#C_2H_6#) undergoes combustion according to the unbalanced equation #C_2H_6+O_2->CO_2+H_2O#,  how much oxygen is required?

Caleb Adams · Accepted Answer

#19.2molO_2# or #615gO_2#

The following is the balanced equation for ethane combustion:

#2C_2H_6+7O_2->4CO_2+6H_2O#

Then number of mole of oxygen that will be needed to completely react with #5.49mol# of ethane could be calculated as follows:

#?molO_2=5.49 cancel(mol C_2H_6)xx(7molO_2)/(2cancel(molC_2H_6))=19.2molO_2#

The required mass of oxygen would be:

#?gO_2=19.2 cancel(molO_2)xx(32.0gO_2)/(1cancel(molO_2))=615gO_2#

Jeremiah Adams · Answer

undefined To balance the combustion equation for ethane (C₂H₆), we first need to determine the stoichiometric coefficients for each compound:

C₂H₆ + O₂ → CO₂ + H₂O

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

According to the balanced equation, for every 2 moles of ethane (C₂H₆) consumed, 7 moles of oxygen (O₂) are required.

Given that you have 5.49 moles of ethane, you can use the stoichiometric ratio to find the amount of oxygen required:

(5.49 moles C₂H₆) × (7 moles O₂ / 2 moles C₂H₆) = 19.215 moles O₂

So, 19.215 moles of oxygen are required for the combustion of 5.49 moles of ethane.

If 5.49 mol of ethane (#C_2H_6#) undergoes combustion according to the unbalanced equation #C_2H_6+O_2-&gt;CO_2+H_2O#, how much oxygen is required?