Question

A sample of propane (C3H8) is placed in a closed vessel together with an amount of O2 that is 2.30 times the amount needed to completely oxidize the propane to CO2 and H2O at constant temperature. Calculate the mole fraction of each component in the resulting mixture after oxidation, assuming that the H2O is present as a gas.Enter your answers to three significant figures separated by commas.

Answer 1

Explanation:

The balanced reaction equation is as follows.

          $C_{3}H_{8} + 5O_{2} \rightarrow 4H_{2}O + 3CO_{2 }$

As it is given that 2.3 times the amount of $O_{2}$ is needed to drive this reaction we can calculate just how much $O_{2}$ we have, keeping in mind we only need 5 moles of $O_{2}$ .

Hence, $5 moles \times 2.30$ = 11.5 moles of $O_{2}$

                              = 12 (moles approx)

Therefore, it means that $C_{3}H{8}$ is our limiting reagent, so product formed will be as follows.

$C_{3}H_{8}$ : $H_{2}O$ = 1:4 meaning we create 4 moles of $H_{2}O$

$C_{3}H_{8}$ : $CO_{2}$ = 1:3 meaning we create 3 moles of $CO_{2}$

As, in the given reaction we had 12 moles of $O_{2}$, but 5 moles were consumed in the reaction meaning we have:

           12 - 5 = 7 moles of $O_{2}$

Thus, we will add up all the moles and get the ratios for each gas as follows.

7 + 4 + 3 = 14 moles

   $O_{2} = \frac{7}{14} = 0.5 [tex]H_{2}O = \frac{4}{14}$ = 0.285

  $CO_{2} = \frac{3}{14}$ = 0.214

Hence, $C_{3}H_{8}$ will be consumed completely.

Answer 2

Answer:

Mol fraction of O2 = 0.482

Mol fraction of H2O = 0.296

Mol fraction of CO2 = 0.222

Explanation:

Step 1: Data given

A sample of propane (C3H8) is placed in a closed vessel together with an amount of O2 that is 2.30 times the amount needed to completely oxidize the propane

Step 2: The balanced equation

C3H8 + 5O2→4H2O + 3CO2

Step 3: Calculate moles

We have 2.30 times the amount of O2 needed to drive this reaction we can calculate just how much O2 we have, keeping in mind we only need 5 moles of O2

Moles O2 = 5 moles * 2.3 = 11.5 moles of O2

⇒ C3H8 is our limiting reagent

Mol ratio C3H8 : H2O = 1:4 meaning for each mol C3H8 consumed, we produce 4 moles H2O

Mol ratio C3H8 : CO2 = 1:3 meaning  for each mol C3H8 consumed, we produce 3 moles CO2

in the reaction we had 11.5 moles of O2, but 5 moles were consumed in the reaction meaning we have:

11.5 - 5 = 6.5 moles of O2  remaining

Step 4: Calculate total moles  

Total moles = 6.5 + 4 + 3 = 13.5moles

Step 5: Calculate mol fraction

O2 =6.5/13.5 = 0.482

H2O = 4/13.5 = 0.296

CO2 = 3/13.5 = 0.222