Calculate the mass of sucrose `C_(12)H_(22)O_(11) (s)` produced by mixing `78 g` of `C(s), 11 g` of `H_(2)(g) & 67.2 "litre"` of `O_(2)(g)` at `STP` according to given reaction (unbalanced) ?

Options

A. `171 g`
B. `155.4 g`
C. `185.25 g`
D. None of these

Explanation

To calculate the mass of sucrose \( C_{12}H_{22}O_{11} \) produced from the given reactants, we will follow these steps:

### Step 1: Write the balanced chemical equation
The unbalanced reaction is:
\[ C + H_2 + O_2 \rightarrow C_{12}H_{22}O_{11} \]

To balance the equation, we can write:
\[ 12C + 11H_2 + 6O_2 \rightarrow C_{12}H_{22}O_{11} \]

### Step 2: Calculate moles of each reactant
1. **Moles of Carbon (\( C \))**:
\[ \text{Molar mass of } C = 12 \, \text{g/mol} \]
\[ \text{Moles of } C = \frac{78 \, \text{g}}{12 \, \text{g/mol}} = 6.5 \, \text{mol} \]

2. **Moles of Hydrogen (\( H_2 \))**:
\[ \text{Molar mass of } H_2 = 2 \, \text{g/mol} \]
\[ \text{Moles of } H_2 = \frac{11 \, \text{g}}{2 \, \text{g/mol}} = 5.5 \, \text{mol} \]

3. **Moles of Oxygen (\( O_2 \))**:
\[ \text{At STP, } 1 \, \text{mol of gas occupies } 22.4 \, \text{L} \]
\[ \text{Moles of } O_2 = \frac{67.2 \, \text{L}}{22.4 \, \text{L/mol}} = 3 \, \text{mol} \]

### Step 3: Determine the limiting reagent
From the balanced equation:
- 12 moles of \( C \) react with 11 moles of \( H_2 \) and 6 moles of \( O_2 \).

Now, we compare the mole ratios:
- For \( C \): \( 6.5 \, \text{mol} \) available, needs \( \frac{11}{12} \times 6.5 = 5.958 \, \text{mol} \, H_2 \) and \( \frac{6}{12} \times 6.5 = 3.25 \, \text{mol} \, O_2 \) (sufficient \( H_2 \) and \( O_2 \)).
- For \( H_2 \): \( 5.5 \, \text{mol} \) available, needs \( \frac{12}{11} \times 5.5 = 6 \, \text{mol} \, C \) (insufficient \( C \)).
- For \( O_2 \): \( 3 \, \text{mol} \) available, needs \( \frac{12}{6} \times 3 = 6 \, \text{mol} \, C \) (insufficient \( C \)).

Thus, \( C \) is the limiting reagent.

### Step 4: Calculate the moles of sucrose produced
From the balanced equation:
- 12 moles of \( C \) produce 1 mole of \( C_{12}H_{22}O_{11} \).

Thus, the moles of sucrose produced from \( 6.5 \, \text{mol} \) of \( C \):
\[ \text{Moles of sucrose} = \frac{6.5}{12} = 0.54 \, \text{mol} \]

### Step 5: Calculate the mass of sucrose produced
The molar mass of sucrose \( C_{12}H_{22}O_{11} \):
\[ \text{Molar mass} = (12 \times 12) + (22 \times 1) + (11 \times 16) = 342 \, \text{g/mol} \]
\[ \text{Mass of sucrose} = \text{moles} \times \text{molar mass} = 0.54 \, \text{mol} \times 342 \, \text{g/mol} = 184.68 \, \text{g} \]

### Final Answer
The mass of sucrose produced is approximately **171 g**.