Westonci.ca is your trusted source for accurate answers to all your questions. Join our community and start learning today! Connect with professionals ready to provide precise answers to your questions on our comprehensive Q&A platform. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.

a. If [tex]4.50 \, \text{mol}[/tex] of ethane, [tex]C_2H_6[/tex], undergoes combustion according to the unbalanced equation [tex]C_2H_6 + O_2 \longrightarrow CO_2 + H_2O[/tex], how many moles of oxygen are required?

b. How many moles of each product are formed?

Sagot :

Sure, I'd be happy to walk you through the steps for solving this problem.

First, it's important to begin with the balanced chemical equation for the combustion of ethane ([tex]\(C_2H_6\)[/tex]). The balanced equation is:

[tex]\[2 C_2H_6 + 7 O_2 \rightarrow 4 CO_2 + 6 H_2O\][/tex]

Given values: We have 4.50 moles of ethane ([tex]\(C_2H_6\)[/tex]).

### (a) Moles of Oxygen Required

From the balanced equation, we know that 2 moles of ethane ([tex]\(C_2H_6\)[/tex]) react with 7 moles of oxygen ([tex]\(O_2\)[/tex]).

To find out how many moles of oxygen are required to combust 4.50 moles of ethane, we use the stoichiometry of the reaction. The relationship can be set up as a ratio:

[tex]\[ \frac{7 \text{ moles of } O_2}{2 \text{ moles of } C_2H_6} \][/tex]

We multiply this ratio by the number of moles of ethane we have:

[tex]\[ \text{moles of } O_2 = \left(\frac{7}{2}\right) \times 4.50 \text{ moles of } C_2H_6 \][/tex]

Carrying out this multiplication:

[tex]\[ \text{moles of } O_2 = 15.75 \text{ moles} \][/tex]

So, 15.75 moles of oxygen are required to combust 4.50 moles of ethane.

### (b) Moles of Each Product Formed

Now, we need to determine how many moles of each product ([tex]\(CO_2\)[/tex] and [tex]\(H_2O\)[/tex]) are formed.

#### Moles of [tex]\(CO_2\)[/tex] Formed

From the balanced equation, 2 moles of [tex]\(C_2H_6\)[/tex] produce 4 moles of [tex]\(CO_2\)[/tex].

Using the ratio:

[tex]\[ \frac{4 \text{ moles of } CO_2}{2 \text{ moles of } C_2H_6} \][/tex]

Multiply this ratio by 4.50 moles of [tex]\(C_2H_6\)[/tex]:

[tex]\[ \text{moles of } CO_2 = \left(\frac{4}{2}\right) \times 4.50 \text{ moles of } C_2H_6 = 9.00 \text{ moles} \][/tex]

So, 9.00 moles of carbon dioxide ([tex]\(CO_2\)[/tex]) are formed.

#### Moles of [tex]\(H_2O\)[/tex] Formed

From the balanced equation, 2 moles of [tex]\(C_2H_6\)[/tex] produce 6 moles of [tex]\(H_2O\)[/tex].

Using the ratio:

[tex]\[ \frac{6 \text{ moles of } H_2O}{2 \text{ moles of } C_2H_6} \][/tex]

Multiply this ratio by 4.50 moles of [tex]\(C_2H_6\)[/tex]:

[tex]\[ \text{moles of } H_2O = \left(\frac{6}{2}\right) \times 4.50 \text{ moles of } C_2H_6 = 13.50 \text{ moles} \][/tex]

So, 13.50 moles of water ([tex]\(H_2O\)[/tex]) are formed.

### Summary:

(a) The number of moles of oxygen required is 15.75 moles.

(b) The number of moles of each product formed is:
- 9.00 moles of carbon dioxide ([tex]\(CO_2\)[/tex])
- 13.50 moles of water ([tex]\(H_2O\)[/tex])