Answer :
To determine how many moles of [tex]\( O_2 \)[/tex] are needed to react with 24 moles of [tex]\( C_2H_6 \)[/tex], we can use the stoichiometric relationship provided by the balanced chemical equation:
[tex]\[ 2 C_2H_6 + 7 O_2 \rightarrow 4 CO_2 + 6 H_2O \][/tex]
Step-by-Step Solution:
1. Identify the given information:
- We have [tex]\( 24 \)[/tex] moles of [tex]\( C_2H_6 \)[/tex].
2. Determine the mole ratio from the balanced equation:
- According to the equation, [tex]\( 2 \)[/tex] moles of [tex]\( C_2H_6 \)[/tex] react with [tex]\( 7 \)[/tex] moles of [tex]\( O_2 \)[/tex].
3. Set up the proportion using the given and required moles:
[tex]\[ \frac{24 \text{ moles of } C_2H_6}{X \text{ moles of } O_2} = \frac{2 \text{ moles of } C_2H_6}{7 \text{ moles of } O_2} \][/tex]
4. Solve for [tex]\( X \)[/tex] which represents the moles of [tex]\( O_2 \)[/tex] needed:
[tex]\[ \frac{24}{X} = \frac{2}{7} \][/tex]
To solve for [tex]\( X \)[/tex], cross-multiply:
[tex]\[ 24 \times 7 = 2 \times X \][/tex]
[tex]\[ 168 = 2X \][/tex]
Now, divide both sides by 2:
[tex]\[ X = \frac{168}{2} \][/tex]
[tex]\[ X = 84 \][/tex]
Thus, the moles of [tex]\( O_2 \)[/tex] needed to react with 24 moles of [tex]\( C_2H_6 \)[/tex] are [tex]\( 84 \)[/tex] moles.
[tex]\[ 2 C_2H_6 + 7 O_2 \rightarrow 4 CO_2 + 6 H_2O \][/tex]
Step-by-Step Solution:
1. Identify the given information:
- We have [tex]\( 24 \)[/tex] moles of [tex]\( C_2H_6 \)[/tex].
2. Determine the mole ratio from the balanced equation:
- According to the equation, [tex]\( 2 \)[/tex] moles of [tex]\( C_2H_6 \)[/tex] react with [tex]\( 7 \)[/tex] moles of [tex]\( O_2 \)[/tex].
3. Set up the proportion using the given and required moles:
[tex]\[ \frac{24 \text{ moles of } C_2H_6}{X \text{ moles of } O_2} = \frac{2 \text{ moles of } C_2H_6}{7 \text{ moles of } O_2} \][/tex]
4. Solve for [tex]\( X \)[/tex] which represents the moles of [tex]\( O_2 \)[/tex] needed:
[tex]\[ \frac{24}{X} = \frac{2}{7} \][/tex]
To solve for [tex]\( X \)[/tex], cross-multiply:
[tex]\[ 24 \times 7 = 2 \times X \][/tex]
[tex]\[ 168 = 2X \][/tex]
Now, divide both sides by 2:
[tex]\[ X = \frac{168}{2} \][/tex]
[tex]\[ X = 84 \][/tex]
Thus, the moles of [tex]\( O_2 \)[/tex] needed to react with 24 moles of [tex]\( C_2H_6 \)[/tex] are [tex]\( 84 \)[/tex] moles.