Urea`(H_(2)NCONH_(2))` is manufactured by passing `CO_(2)`(g) through ammonia solution followed by crystallization. For the above reaction is prepared by combustion of hydrocarbons. If combustion of 236 kg of a saturated hydrocarbon `(C_(n)H_(2n+2))` produces as much `CO_(2)` as required for production of 999.6 kg urea then molecular formula of hydrocarbon is:

Options

A. `C_(10)H_(22)`
B. `C_(12)H_(26)`
C. `C_(13)H_(28)`
D. `C_(8)H_(18)`

Answer

To solve the problem, we need to determine the molecular formula of the saturated hydrocarbon \( C_nH_{2n+2} \) that produces the same amount of \( CO_2 \) as is required for the production of 999.6 kg of urea. Here’s a step-by-step breakdown of the solution:

### Step 1: Write the combustion reaction of the hydrocarbon
The combustion of a saturated hydrocarbon \( C_nH_{2n+2} \) can be represented as:
\[ C_nH_{2n+2} + O_2 \rightarrow nCO_2 + (n+1)H_2O \]
From this reaction, we see that 1 mole of the hydrocarbon produces \( n \) moles of \( CO_2 \).

### Step 2: Calculate the molar mass of the hydrocarbon
The molar mass of the hydrocarbon \( C_nH_{2n+2} \) is calculated as follows:
\[ \text{Molar mass} = 12n + (2n + 2) = 14n + 2 \text{ g/mol} \]

### Step 3: Determine the moles of hydrocarbon in 236 kg
Given that 236 kg of the hydrocarbon is burned, we convert this to grams:
\[ 236 \text{ kg} = 236 \times 10^3 \text{ g} \]
The number of moles of hydrocarbon is:
\[ \text{Moles of hydrocarbon} = \frac{236 \times 10^3 \text{ g}}{14n + 2} \]

### Step 4: Calculate the moles of \( CO_2 \) produced
Since 1 mole of hydrocarbon produces \( n \) moles of \( CO_2 \):
\[ \text{Moles of } CO_2 = n \times \frac{236 \times 10^3}{14n + 2} \]

### Step 5: Relate \( CO_2 \) to urea production
From the reaction of \( CO_2 \) with ammonia to form urea:
\[ CO_2 + 2NH_3 \rightarrow NH_2CONH_2 + H_2O \]
1 mole of \( CO_2 \) produces 1 mole of urea. The molar mass of urea is 60 g/mol. Thus, the moles of urea produced from 999.6 kg is:
\[ \text{Moles of urea} = \frac{999.6 \times 10^3 \text{ g}}{60 \text{ g/mol}} = 16660 \text{ moles} \]

### Step 6: Set the moles of \( CO_2 \) equal to the moles of urea
Since the moles of \( CO_2 \) produced equals the moles of urea:
\[ n \times \frac{236 \times 10^3}{14n + 2} = 16660 \]

### Step 7: Solve for \( n \)
Rearranging gives:
\[ n \times 236000 = 16660 \times (14n + 2) \]
Expanding and simplifying:
\[ 236000n = 233240n + 33320 \]
\[ 236000n - 233240n = 33320 \]
\[ 2760n = 33320 \]
\[ n = \frac{33320}{2760} \approx 12.072 \approx 12 \]

### Step 8: Determine the molecular formula
Substituting \( n = 12 \) into the formula \( C_nH_{2n+2} \):
\[ C_{12}H_{2(12)+2} = C_{12}H_{26} \]

### Final Answer
The molecular formula of the saturated hydrocarbon is:
\[ \boxed{C_{12}H_{26}} \]