For reaction `A+2BtoC`. The amount of C formed by starting the reaction with 5 mole of A and 8 mole of B is :

Options

A. 5 mol
B. 8 mol
C. 16 mol
D. 4 mol

Answer

To determine the amount of C formed in the reaction \( A + 2B \rightarrow C \) when starting with 5 moles of A and 8 moles of B, we need to identify the limiting reagent and then calculate the amount of product formed based on that.

### Step-by-Step Solution:

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

**Step 2: Identify the stoichiometric ratios.**
From the balanced equation, we see that:
- 1 mole of A reacts with 2 moles of B to produce 1 mole of C.

**Step 3: Determine the amount of B needed for the available moles of A.**
Given that we have 5 moles of A, we can calculate the moles of B required:
- For 5 moles of A, the moles of B required = \( 5 \text{ moles A} \times 2 \text{ moles B/mole A} = 10 \text{ moles B} \).

**Step 4: Compare the required moles of B with the available moles of B.**
We have only 8 moles of B available, but we need 10 moles of B to react with all 5 moles of A. This indicates that B is the limiting reagent.

**Step 5: Calculate the amount of C produced based on the limiting reagent (B).**
Since B is the limiting reagent, we will calculate how much C can be formed from the available moles of B:
- From the balanced equation, 2 moles of B produce 1 mole of C. Therefore, for 8 moles of B:
\[ \text{Moles of C produced} = \frac{8 \text{ moles B}}{2 \text{ moles B/mole C}} = 4 \text{ moles C} \]

### Final Answer:
The amount of C formed is 4 moles.

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