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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])
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])
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