Question

How many liters of CO2 are produced from the reaction of 2.5 L of propane (C3H8) with 12 L of O2, as shown below: C3H8 (g) + 5 O2 (g) → 3 CO2 (g) + 4 H2O (g)

Answer 1

We can use stoichiometry to determine the amount of CO2 produced from the given reaction. The balanced chemical equation tells us that 1 mole of C3H8 reacts with 5 moles of O2 to produce 3 moles of CO2 and 4 moles of H2O.

First, we need to determine the number of moles of propane (C3H8) and oxygen (O2) present in the reaction mixture:

- Number of moles of C3H8 = (volume of C3H8 at STP) / (molar volume of C3H8 at STP)

= (2.5 L) / (22.4 L/mol)

= 0.111 moles of C3H8

- Number of moles of O2 = (volume of O2 at STP) / (molar volume of O2 at STP)

= (12 L) / (22.4 L/mol)

= 0.536 moles of O2

Next, we need to determine the limiting reactant. The reactant that is completely consumed is the limiting reactant and determines the maximum amount of product formed. To find the limiting reactant, we compare the moles of each reactant to their respective stoichiometric coefficients in the balanced chemical equation:

- For C3H8: 0.111 moles x (5 mol O2 / 1 mol C3H8) = 0.556 mol O2 required

- For O2: 0.536 moles present

Since the amount of O2 required (0.556 mol) is greater than the amount of O2 present (0.536 mol), O2 is the limiting reactant.

The balanced chemical equation tells us that for every 5 moles of O2, 3 moles of CO2 are produced. So we can set up a ratio to determine the moles of CO2 produced:

- (0.536 mol O2) x (3 mol CO2 / 5 mol O2) = 0.322 mol CO2

Finally, we can convert moles of CO2 to volume using the ideal gas law at STP:

- PV = nRT, where P = pressure (1 atm), V = volume (unknown), n = moles (0.322 mol), R = gas constant (0.08206 L atm mol^-1 K^-1), T = temperature (273 K)

- Solving for V: V = nRT / P = (0.322 mol)(0.08206 L atm mol^-1 K^-1)(273 K) / (1 atm) = 7.36 L

- Therefore, the reaction of 2.5 L of propane (C3H8) with 12 L of O2 produces 7.36 L of CO2 at STP.

Answer 2

Therefore, from the given amounts of propane and oxygen, 7.50 liters of CO2 are produced in the reaction.

Step-by-step explanation:

To find the volume of CO2 produced from the reaction of 2.5 L of propane (C3H8) with 12 L of O2, you can use the balanced equation:

C3H8 (g)

+

5

O2 (g)

→

3

CO2 (g)

+

4

H2O (g)

C3H8 (g)+5O2 (g)→3CO2 (g)+4H2O (g)

The balanced equation shows that 1 mole of C3H8 produces 3 moles of CO2.

First, let's determine the number of moles of propane (C3H8) and oxygen (O2) provided:

Given:

Volume of C3H8 = 2.5 L

Volume of O2 = 12 L

The balanced equation ratio between C3H8 and O2 is 1:5. Therefore, the limiting reactant is C3H8 because there is less propane than required for the reaction with the available oxygen.

Convert the volumes of C3H8 and O2 to moles using the ideal gas law (assuming standard conditions):

moles

=

volume (L)

molar volume at STP (22.4 L/mol)

moles=

molar volume at STP (22.4 L/mol)

volume (L)

​

Moles of C3H8:

moles of C3H8

=

2.5

 

L

22.4

 

L/mol

moles of C3H8=

22.4L/mol

2.5L

​

Moles of O2:

moles of O2

=

12

 

L

22.4

 

L/mol

moles of O2=

22.4L/mol

12L

​

Determine the limiting reactant by comparing the moles of C3H8 and O2 using the balanced equation ratio.

Moles of CO2 produced will be based on the moles of C3H8 used (since it's the limiting reactant) using the stoichiometry in the balanced equation.

Finally, convert the moles of CO2 to liters using the molar volume of a gas at standard conditions (22.4 L/mol).Calculate the moles of C3H8 and O2:

Moles of C3H8

=

2.5

 

L

22.4

 

L/mol

=

0.1116

 

moles

Moles of C3H8=

22.4L/mol

2.5L

​

=0.1116moles

Moles of O2

=

12

 

L

22.4

 

L/mol

=

0.5357

 

moles

Moles of O2=

22.4L/mol

12L

​

=0.5357moles

Determine the limiting reactant:

The balanced equation ratio between C3H8 and O2 is 1:5. There is less C3H8 (0.1116 moles) compared to the required amount based on the ratio (0.1116 * 5 = 0.558 moles needed), indicating that C3H8 is the limiting reactant.

Calculate moles of CO2 produced from the moles of C3H8 used:

Since 1 mole of C3H8 produces 3 moles of CO2 according to the balanced equation:

Moles of CO2

=

0.1116

 

moles of C3H8

×

3

 

moles of CO2

1

 

mole of C3H8

=

0.3348

 

moles of CO2

Moles of CO2=0.1116moles of C3H8×

1mole of C3H8

3moles of CO2

​

=0.3348moles of CO2

Convert moles of CO2 to liters:

Volume of CO2

=

Moles of CO2

×

Molar volume at STP

=

0.3348

 

moles

×

22.4

 

L/mol

=

7.50

 

Liters

Volume of CO2=Moles of CO2×Molar volume at STP=0.3348moles×22.4L/mol=7.50Liters

Therefore, from the given amounts of propane and oxygen, 7.50 liters of CO2 are produced in the reaction.