9 moles of "D" and 14 moles of E are allowed to react in a closed vessel according to given reactions. Calculate number of moles of B formed in the end of reaction, if 4 moles of G are present in reaction vessel. (percentage yield of reaction is mentioned in the reaction) Step -1 `3D+4E 80%rarr5C+A` Step-2 `3C+5G 50%rarr 6B+F`

Options

A. `2.4`
B. `30`
C. `4.8`
D. `1`

Answer

To solve the problem step by step, we will follow the reactions provided, determine the limiting reagents, and calculate the number of moles of B formed at the end of the reactions.

### Step 1: Identify the reactions
We have two reactions:
1. \( 3D + 4E \xrightarrow{80\%} 5C + A \)
2. \( 3C + 5G \xrightarrow{50\%} 6B + F \)

### Step 2: Determine the limiting reagent for the first reaction
We have:
- Moles of D = 9
- Moles of E = 14

From the first reaction, the stoichiometry is:
- 3 moles of D react with 4 moles of E.

To find the limiting reagent, we calculate how many moles of D and E are needed:
- For 9 moles of D:
\[ \text{Moles of E required} = \frac{4}{3} \times 9 = 12 \text{ moles of E} \]
- For 14 moles of E:
\[ \text{Moles of D required} = \frac{3}{4} \times 14 = 10.5 \text{ moles of D} \]

Now we compare:
- D requires 12 moles of E (we have 14, so sufficient).
- E requires 10.5 moles of D (we have 9, so insufficient).

**Conclusion**: D is the limiting reagent.

### Step 3: Calculate moles of C produced
From the balanced equation:
- 3 moles of D produce 5 moles of C.

Using the amount of D we have:
\[ \text{Moles of C produced} = \frac{5}{3} \times 9 = 15 \text{ moles of C} \]

Considering the percentage yield of 80%:
\[ \text{Actual moles of C} = 15 \times \frac{80}{100} = 12 \text{ moles of C} \]

### Step 4: Determine the limiting reagent for the second reaction
Now we have:
- Moles of C = 12
- Moles of G = 4

From the second reaction:
- 3 moles of C react with 5 moles of G.

Calculating the required moles:
- For 12 moles of C:
\[ \text{Moles of G required} = \frac{5}{3} \times 12 = 20 \text{ moles of G} \]
- For 4 moles of G:
\[ \text{Moles of C required} = \frac{3}{5} \times 4 = 2.4 \text{ moles of C} \]

**Conclusion**: G is the limiting reagent.

### Step 5: Calculate moles of B produced
From the balanced equation:
- 5 moles of G produce 6 moles of B.

Using the amount of G we have:
\[ \text{Moles of B produced} = \frac{6}{5} \times 4 = 4.8 \text{ moles of B} \]

Considering the percentage yield of 50%:
\[ \text{Actual moles of B} = 4.8 \times \frac{50}{100} = 2.4 \text{ moles of B} \]

### Final Answer
The number of moles of B formed at the end of the reaction is **2.4 moles**.

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