Sure, let's work through each part of the question step by step!
### Problem:
Given the reaction:
[tex]\[ 2 \text{C}_2\text{H}_6 + 7 \text{O}_2 \rightarrow 4 \text{CO}_2 + 6 \text{H}_2\text{O} \][/tex]
#### 1.1 How many moles of CO[tex]\(_2\)[/tex] are produced when B moles of C[tex]\(_2\)[/tex]H[tex]\(_6\)[/tex] are used up?
From the balanced equation, we notice that 2 moles of C[tex]\(_2\)[/tex]H[tex]\(_6\)[/tex] produce 4 moles of CO[tex]\(_2\)[/tex]. This means that each mole of C[tex]\(_2\)[/tex]H[tex]\(_6\)[/tex] produces twice the number of moles of CO[tex]\(_2\)[/tex].
So, if we have B moles of C[tex]\(_2\)[/tex]H[tex]\(_6\)[/tex]:
[tex]\[ \text{Moles of CO}_2 = B \times 2 \][/tex]
Given B = 1 for demonstration:
[tex]\[ \text{Moles of CO}_2 = 1 \times 2 = 2 \][/tex]
#### 1.2 How many moles of H[tex]\(_2\)[/tex]O are needed to produce 18 moles of CO[tex]\(_2\)[/tex]?
From the balanced equation, we see that 4 moles of CO[tex]\(_2\)[/tex] correspond to 6 moles of H[tex]\(_2\)[/tex]O. To find the ratio:
[tex]\[ \frac{6 \text{ moles H}_2\text{O}}{4 \text{ moles CO}_2} \][/tex]
Thus, for 18 moles of CO[tex]\(_2\)[/tex]:
[tex]\[ \text{Moles of H}_2\text{O} = 18 \times \frac{6}{4} = 18 \times 1.5 = 27 \][/tex]
[tex]\[ \][/tex]
#### 1.3 How many moles of O[tex]\(_2\)[/tex] react with 6 moles of C[tex]\(_2\)[/tex]H[tex]\(_6\)[/tex]?
From the balanced equation, we see that 2 moles of C[tex]\(_2\)[/tex]H[tex]\(_6\)[/tex] react with 7 moles of O[tex]\(_2\)[/tex]. To determine how much O[tex]\(_2\)[/tex] would be required for 6 moles of C[tex]\(_2\)[/tex]H[tex]\(_6\)[/tex]:
[tex]\[ \frac{7 \text{ moles O}_2}{2 \text{ moles C}_2\text{H}_6} \][/tex]
Thus, for 6 moles of C[tex]\(_2\)[/tex]H[tex]\(_6\)[/tex]:
[tex]\[ \text{Moles of O}_2 = 6 \times \frac{7}{2} = 6 \times 3.5 = 21 \][/tex]
### Summary
1. 2 moles of CO[tex]\(_2\)[/tex] are produced when 1 mole of C[tex]\(_2\)[/tex]H[tex]\(_6\)[/tex] is used up.
2. 27 moles of H[tex]\(_2\)[/tex]O are needed to produce 18 moles of CO[tex]\(_2\)[/tex].
3. 21 moles of O[tex]\(_2\)[/tex] react with 6 moles of C[tex]\(_2\)[/tex]H[tex]\(_6\)[/tex].
