1. Let's start by analyzing the given reaction:
[tex]\[ 2 C_2H_6 + 7 O_2 \rightarrow 4 CO_2 + 6 H_2O \][/tex]
This reaction tells us the stoichiometric ratios between the reactants and the products.
### 1.1 How many moles of [tex]\( CO_2 \)[/tex] are produced when 13 moles of [tex]\( C_2H_6 \)[/tex] are used up?
From the balanced chemical equation:
- 2 moles of [tex]\( C_2H_6 \)[/tex] produce 4 moles of [tex]\( CO_2 \)[/tex].
To find out how many moles of [tex]\( CO_2 \)[/tex] are produced by 13 moles of [tex]\( C_2H_6 \)[/tex], we can set up the following ratio:
[tex]\[
\frac{4 \text{ moles } CO_2}{2 \text{ moles } C_2H_6} = \frac{x \text{ moles } CO_2}{13 \text{ moles } C_2H_6}
\][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[
x = 13 \times \frac{4}{2}
\][/tex]
[tex]\[
x = 13 \times 2
\][/tex]
[tex]\[
x = 26
\][/tex]
So, 26 moles of [tex]\( CO_2 \)[/tex] are produced when 13 moles of [tex]\( C_2H_6 \)[/tex] are used up.
### 1.2 How many moles of [tex]\( H_2O \)[/tex] are needed to produce 18 moles of [tex]\( CO_2 \)[/tex]?
From the balanced chemical equation:
- 4 moles of [tex]\( CO_2 \)[/tex] are produced along with 6 moles of [tex]\( H_2O \)[/tex].
To find out how many moles of [tex]\( H_2O \)[/tex] are needed to produce 18 moles of [tex]\( CO_2 \)[/tex], we can set up the following ratio:
[tex]\[
\frac{6 \text{ moles } H_2O}{4 \text{ moles } CO_2} = \frac{y \text{ moles } H_2O}{18 \text{ moles } CO_2}
\][/tex]
Solving for [tex]\( y \)[/tex]:
[tex]\[
y = 18 \times \frac{6}{4}
\][/tex]
[tex]\[
y = 18 \times 1.5
\][/tex]
[tex]\[
y = 27
\][/tex]
So, 27 moles of [tex]\( H_2O \)[/tex] are needed to produce 18 moles of [tex]\( CO_2 \)[/tex].
### 1.3 How many moles of [tex]\( O_2 \)[/tex] react with 6 moles of [tex]\( C_2H_6 \)[/tex]?
From the balanced chemical equation:
- 2 moles of [tex]\( C_2H_6 \)[/tex] react with 7 moles of [tex]\( O_2 \)[/tex].
To find out how many moles of [tex]\( O_2 \)[/tex] react with 6 moles of [tex]\( C_2H_6 \)[/tex], we can set up the following ratio:
[tex]\[
\frac{7 \text{ moles } O_2}{2 \text{ moles } C_2H_6} = \frac{z \text{ moles } O_2}{6 \text{ moles } C_2H_6}
\][/tex]
Solving for [tex]\( z \)[/tex]:
[tex]\[
z = 6 \times \frac{7}{2}
\][/tex]
[tex]\[
z = 6 \times 3.5
\][/tex]
[tex]\[
z = 21
\][/tex]
So, 21 moles of [tex]\( O_2 \)[/tex] react with 6 moles of [tex]\( C_2H_6 \)[/tex].
