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1. In the reaction [tex]\( 2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O \)[/tex]:

1.1. How many moles of [tex]\( CO_2 \)[/tex] are produced when 1 mole of [tex]\( C_2H_6 \)[/tex] is used up?

1.2. How many moles of [tex]\( H_2O \)[/tex] are produced when 18 moles of [tex]\( CO_2 \)[/tex] are produced?

1.3. How many moles of [tex]\( O_2 \)[/tex] react with 6 moles of [tex]\( C_2H_6 \)[/tex]?

Sagot :

GinnyAnsweridn
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].
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