If 5.27 mol of ethane (C2H6) undergo combus-tion according to the unbalanced equation C2H6 + O2 −! CO2 + H2O, how much oxygen is required?

Answer in units of mol.

Thanks!

1. Write the equation and balance it.

2. Using the coefficients in the balanced equation, convert 5.27 moles ethane to moles oxygen.
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15.81

If 5.27 mol of ethane (C2H6) undergo combustion

according to the unbalanced equation
C2H6 + O2 −→ CO2 + H2O ,
how much oxygen is required?
Answer in units of mol.

To determine how much oxygen is required for the combustion of 5.27 mol of ethane (C2H6), we need to balance the chemical equation first.

The unbalanced equation is: C2H6 + O2 → CO2 + H2O

To balance the equation, we start by counting the number of atoms of each element on both sides of the equation.

On the left side of the equation (reactants):
- There are 2 carbon (C) atoms
- There are 6 hydrogen (H) atoms
- There are 2 oxygen (O) atoms (from the O2 molecule)

On the right side of the equation (products):
- There are 1 carbon (C) atom (from the CO2 molecule)
- There are 2 hydrogen (H) atoms (from the H2O molecule)
- There are 3 oxygen (O) atoms (from the CO2 and H2O molecules)

To balance the equation, we need to ensure that the number of atoms of each element is the same on both sides.

To balance the carbon (C) atoms:
- We multiply the CO2 molecule on the right side by 2: C2H6 + O2 → 2CO2 + H2O

To balance the hydrogen (H) atoms:
- We multiply the H2O molecule on the right side by 3: C2H6 + O2 → 2CO2 + 3H2O

Now, let's count the number of oxygen (O) atoms on both sides:
- On the left side, there are 2 oxygen (O) atoms from the O2 molecule.
- On the right side, there are 4 oxygen (O) atoms from the 2 CO2 molecules, and 3 oxygen (O) atoms from the 3 H2O molecules. In total, there are 7 oxygen (O) atoms on the right side.

To balance the oxygen (O) atoms:
- We multiply the O2 molecule on the left side by 7/2 (3.5): C2H6 + 3.5O2 → 2CO2 + 3H2O

Now that the equation is balanced, we see that 3.5 moles of O2 are required for every mole of ethane (C2H6).

To find the amount of oxygen required for 5.27 mol of ethane, we can multiply the number of moles of ethane by the stoichiometric coefficient of O2:
Oxygen required = 5.27 mol C2H6 * (3.5 mol O2 / 1 mol C2H6)

Calculating the value:
Oxygen required = 5.27 mol C2H6 * 3.5 mol O2 / 1 mol C2H6
Oxygen required = 18.445 mol O2

Therefore, 18.445 mol of oxygen is required for the combustion of 5.27 mol of ethane.