The number of moles of C and D produced on mixing 5 moles of A and 7 moles of B are respectively:
`(3A + 5B rArr 7C + 9D)`

Options

A. 9 moles and 11 moles
B. 11.66 moles and 15 moles
C. 9.8 moles and 12.6 moles
D. 1 mole and 13 moles

Answer

To solve the problem, we need to determine the number of moles of C and D produced when 5 moles of A and 7 moles of B are mixed, based on the given reaction:

\[ 3A + 5B \rightarrow 7C + 9D \]

### Step 1: Identify the limiting reagent
First, we need to find out which reactant is the limiting reagent. The stoichiometry of the reaction tells us that:
- 3 moles of A react with 5 moles of B.

To find out how much A is needed for 7 moles of B:
- From the reaction, 5 moles of B require 3 moles of A.
- Therefore, 1 mole of B requires \( \frac{3}{5} \) moles of A.
- For 7 moles of B, the amount of A required is:

\[ \text{A required} = 7 \times \frac{3}{5} = \frac{21}{5} = 4.2 \text{ moles of A} \]

Since we have 5 moles of A available, A is in excess, and B is the limiting reagent.

### Step 2: Calculate the moles of C produced
Next, we calculate how many moles of C are produced from the limiting reagent (B):
- According to the reaction, 5 moles of B produce 7 moles of C.
- Therefore, for 7 moles of B, the amount of C produced is:

\[ \text{C produced} = 7 \times \frac{7}{5} = \frac{49}{5} = 9.8 \text{ moles of C} \]

### Step 3: Calculate the moles of D produced
Now, we calculate how many moles of D are produced from the limiting reagent (B):
- According to the reaction, 5 moles of B produce 9 moles of D.
- Therefore, for 7 moles of B, the amount of D produced is:

\[ \text{D produced} = 7 \times \frac{9}{5} = \frac{63}{5} = 12.6 \text{ moles of D} \]

### Final Answer
Thus, the number of moles of C and D produced are:
- Moles of C = 9.8
- Moles of D = 12.6