Chemistry - Eudiometry Question with Solution | TestHub

Question Explanation: Calculate the difference in volume of H₂ and C₅H₈ in the original mixture.

Concept: Eudiometry.

Solution:

Let the volume of C₅H₈ = $v$.

The volume of H₂ = $40-v$.

 

The reaction is:

C₅H₈ + 2H₂ → C₅H₁₂

Initial: $v$ $40-v$

After reaction: $v-x$ $40-v-2x$ $x$

Given that

$$v_i-v_f=24$$

 

 

$40-(v-x+40-v-2x+x)=24$

$2x=24$

 

$x=12ml$

 

So,

$$C_5{H}_8+2H_2\to C_5{H}_{12}$$

 

Initial $v-x$

$40-v-2x$

$x$

later $v-12$

$16-v$

12

Now

LR excess

C₅H₁₂ + 8O₂ → 5CO₂ + 6H₂O

 

12 V₁

- V₁-96 60 X

 

C₅H₈ + 7O₂ → 5CO₂ + 4H₂O (I)

V-12 V₂

- V₂-7(V-12) 5(V-12) X

 

 

 

H₂ + $\frac{1}{2}$O₂ → H₂O

16-v V₃

- V₃-1/2[16-V] X

 

Now, there is a contraction of volume by 60 ml after combustion.

$$v_i-v_f=60$$

 

$\left[\left(12+v_1\right)\right]+(v-12+v_2)+\left[(16-v+v_3)]-\left[v_1-96+60\right)+\left(v_2-7\left(v-12\right)\right]\right]$

$+5\left(v-12\right)+v_3-\frac{1}{2}\left(16-v\right)=60$

$\left(v_1+v_{2+}{v}_3+16\right)-\left(v_1+v_2+v_3-\frac{3v}{2}-20\right)=60$

$$\begin{gathered} \text{displaylines}16+\frac{3v}{2}+20=60 \\ \frac{3v}{2}=24 \\ v=16ml \end{gathered}$$

So, the initial mixture is:

C₅H₈ + H₂

 

C₅H₈ = $v=16$ ml

H₂ = $40-v=40-16=24$ ml

 

Difference in volume of H₂ and O₂= $24-16=8$ ml.

 

Final Answer: 8