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 solve the problem of how much C is formed from the reaction \( A + 2B \rightarrow C \) when starting with 5 moles of A and 8 moles of B, we can follow these steps:

### Step 1: Write down 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 limiting reagent.
To find out which reactant is the limiting reagent, we need to compare the available moles of A and B with their stoichiometric requirements:
- For 5 moles of A, the amount of B required would be:
\[ 5 \, \text{moles of A} \times 2 \, \text{moles of B/mole of A} = 10 \, \text{moles of B} \]

- However, we only have 8 moles of B available. Therefore, B is the limiting reagent.

### Step 4: Calculate the amount of C produced based on the limiting reagent.
Since B is the limiting reagent, we will calculate how much C can be produced from the available 8 moles of B:
- According to the stoichiometry of the reaction, 2 moles of B produce 1 mole of C. Thus, from 8 moles of B, we can calculate the amount of C produced:
\[ \text{Moles of C} = \frac{8 \, \text{moles of B}}{2 \, \text{moles of B/mole of C}} = 4 \, \text{moles of C} \]

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

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