14 moles of A and 20 moles of B, 8 moles of C is formed in the following reaction `3A+4Brarr2C` then.

Options

A. 8 mole of B react
B. 10 mole of A react
C. 4 mole of B left
D. 8 mole of A left.

Explanation

To solve the problem, we need to analyze the reaction and determine how many moles of reactants A and B are left after the formation of 8 moles of product C.

### Step-by-Step Solution:

1. **Write the Balanced Reaction:**
The balanced chemical equation is:
\[ 3A + 4B \rightarrow 2C \]

2. **Identify Initial Moles:**
We start with:
- Moles of A = 14
- Moles of B = 20

3. **Determine Moles of C Formed:**
According to the problem, 8 moles of C are formed.

4. **Use Stoichiometry to Find Moles of A and B Consumed:**
From the balanced equation, we see that:
- 2 moles of C are produced from 3 moles of A and 4 moles of B.
- Therefore, for 8 moles of C produced, we can set up a proportion to find out how many moles of A and B are consumed.

For C:
\[ \text{Moles of A consumed} = \left( \frac{3 \text{ moles of A}}{2 \text{ moles of C}} \right) \times 8 \text{ moles of C} = 12 \text{ moles of A} \]
\[ \text{Moles of B consumed} = \left( \frac{4 \text{ moles of B}}{2 \text{ moles of C}} \right) \times 8 \text{ moles of C} = 16 \text{ moles of B} \]

5. **Calculate Remaining Moles of A and B:**
- Remaining moles of A:
\[ \text{Remaining A} = \text{Initial A} - \text{Consumed A} = 14 - 12 = 2 \text{ moles of A} \]
- Remaining moles of B:
\[ \text{Remaining B} = \text{Initial B} - \text{Consumed B} = 20 - 16 = 4 \text{ moles of B} \]

6. **Conclusion:**
After the reaction, we have:
- 2 moles of A left
- 4 moles of B left

### Final Answer:
The correct option is that there are 2 moles of A left and 4 moles of B left.

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