For the reaction `A + 2B to C`, 5 moles of A and 8 mole of B will produce

Answer

To solve the problem of how many moles of C will be produced from the reaction \( A + 2B \rightarrow C \) given 5 moles of A and 8 moles of B, we can follow these steps:

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

### Step 2: Determine the mole ratio from the balanced equation.
From the balanced equation, we can see that:
- 1 mole of A reacts with 2 moles of B to produce 1 mole of C.

### Step 3: Calculate the moles of B required for the given moles of A.
To find out how many moles of B are needed to completely react with 5 moles of A:
- Since 1 mole of A requires 2 moles of B, then:
\[ 5 \, \text{moles of A} \times 2 \, \text{moles of B/mole of A} = 10 \, \text{moles of B} \]

### Step 4: Identify the limiting reagent.
We have 8 moles of B available, but we need 10 moles of B to react with 5 moles of A. Therefore, B is the limiting reagent because we do not have enough B to react with all of A.

### Step 5: Calculate the amount of C produced based on the limiting reagent.
According to the balanced equation, 2 moles of B produce 1 mole of C. Therefore, we can calculate how many moles of C can be produced from the available 8 moles of B:
- From the ratio, we can derive:
\[ 8 \, \text{moles of B} \times \frac{1 \, \text{mole of C}}{2 \, \text{moles of B}} = 4 \, \text{moles of C} \]

### Conclusion:
Thus, when 5 moles of A react with 8 moles of B, they will produce **4 moles of C**.

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