In the reaction: `overset(3)(A)+overset(2)(2B)++overset(1)(3C)toD` …. Initial no. of moles
the limiting reactant may be

Options

A. A
B. B
C. C
D. D

Answer

To determine the limiting reactant in the reaction \( A + 2B + 3C \rightarrow D \) with initial moles given as \( A = 3 \), \( B = 2 \), and \( C = 1 \), we can follow these steps:

### Step 1: Write the balanced chemical equation
The balanced chemical equation is:
\[ A + 2B + 3C \rightarrow D \]

### Step 2: Identify the initial number of moles
From the problem, we have:
- Moles of \( A = 3 \)
- Moles of \( B = 2 \)
- Moles of \( C = 1 \)

### Step 3: Determine the stoichiometric coefficients
From the balanced equation:
- The stoichiometric coefficient of \( A \) is 1
- The stoichiometric coefficient of \( B \) is 2
- The stoichiometric coefficient of \( C \) is 3

### Step 4: Calculate the ratio of moles to stoichiometric coefficients
Now, we will calculate the ratio of the initial moles to their respective stoichiometric coefficients:

1. For \( A \):
\[ \text{Ratio for } A = \frac{\text{Moles of } A}{\text{Stoichiometric coefficient of } A} = \frac{3}{1} = 3 \]

2. For \( B \):
\[ \text{Ratio for } B = \frac{\text{Moles of } B}{\text{Stoichiometric coefficient of } B} = \frac{2}{2} = 1 \]

3. For \( C \):
\[ \text{Ratio for } C = \frac{\text{Moles of } C}{\text{Stoichiometric coefficient of } C} = \frac{1}{3} \approx 0.33 \]

### Step 5: Identify the limiting reactant
The limiting reactant is the one with the smallest ratio calculated in the previous step.

- The ratios we calculated are:
- For \( A \): 3
- For \( B \): 1
- For \( C \): 0.33

Since \( C \) has the smallest ratio (0.33), it is the limiting reactant.

### Conclusion
Thus, the limiting reactant in this reaction is \( C \).