Maximum mass of sucrose `C_(12)H_(22)O_(11)` produced by mixing
84 gm of carbon , 12 gm of hydrogen and 56 lit. `O_(2)` at 1 atm & 273K according to given reaction, is
C(s) + `H_(2)`(g) + `O_(2)` (g) `to` `C_(12)H_(22)O_(11)`(s)

Options

A. 138.5
B. 155.5
C. 172.5
D. 199.5

Answer

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

### Step 1: Write and Balance the Reaction
The unbalanced reaction is:
\[ C(s) + H_2(g) + O_2(g) \rightarrow C_{12}H_{22}O_{11}(s) \]

To balance the reaction, we need to ensure that the number of each type of atom is the same on both sides.

After balancing, the reaction becomes:
\[ 12C(s) + 11H_2(g) + 6O_2(g) \rightarrow C_{12}H_{22}O_{11}(s) \]

### Step 2: Calculate Moles of Each Reactant
Now, we will calculate the moles of each reactant provided:

1. **Moles of Carbon (C)**:
\[ \text{Moles of } C = \frac{\text{mass}}{\text{molar mass}} = \frac{84 \text{ g}}{12 \text{ g/mol}} = 7 \text{ moles} \]

2. **Moles of Hydrogen (H2)**:
\[ \text{Moles of } H_2 = \frac{12 \text{ g}}{2 \text{ g/mol}} = 6 \text{ moles} \]

3. **Moles of Oxygen (O2)**:
Given that 1 mole of gas occupies 22.4 L at STP, we can calculate the moles of \( O_2 \):
\[ \text{Moles of } O_2 = \frac{56 \text{ L}}{22.4 \text{ L/mol}} \approx 2.5 \text{ moles} \]

### Step 3: Identify the Limiting Reagent
To find the limiting reagent, we compare the mole ratios from the balanced equation:
- From the balanced equation, 12 moles of \( C \) react with 11 moles of \( H_2 \) and 6 moles of \( O_2 \).
- We have:
- 7 moles of \( C \)
- 6 moles of \( H_2 \)
- 2.5 moles of \( O_2 \)

Since \( O_2 \) is the limiting reagent (as it has the smallest number of moles relative to the stoichiometric requirements), we will use it to calculate the amount of sucrose produced.

### Step 4: Calculate Moles of Sucrose Produced
From the balanced equation:
- 6 moles of \( O_2 \) produce 2 moles of \( C_{12}H_{22}O_{11} \).

Thus, for 1 mole of \( O_2 \):
\[ \text{Moles of } C_{12}H_{22}O_{11} = \frac{2}{6} = \frac{1}{3} \text{ moles} \]

For 2.5 moles of \( O_2 \):
\[ \text{Moles of } C_{12}H_{22}O_{11} = \frac{1}{3} \times 2.5 = \frac{2.5}{3} \approx 0.833 \text{ moles} \]

### Step 5: Calculate the Mass of Sucrose Produced
Now, we calculate the mass of sucrose produced using its molar mass:
- The molar mass of \( C_{12}H_{22}O_{11} \) is calculated as follows:
\[ 12 \times 12 + 22 \times 1 + 11 \times 16 = 144 + 22 + 176 = 342 \text{ g/mol} \]

Thus, the mass of sucrose produced is:
\[ \text{Mass} = \text{Moles} \times \text{Molar Mass} = 0.833 \text{ moles} \times 342 \text{ g/mol} \approx 285.5 \text{ g} \]

### Final Answer
The maximum mass of sucrose produced is approximately **285.5 grams**.

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