In a clinical laborator, a sample of urine containing 0.120 g of urea `NH_(2)CONH_(2)` (M.Wt.60. was treated with excess of nitrous acid. The urea reacted according to the following equation `NH_(2)CONH_(2)+2HNO_(2)rarr CO_(2)+2N_(2)+3H_(2)O`. The gas formed was passed through aqueous sodium hydroxide the final volume is measured at STP, whatwas this volume.

Options

A. 89.6cc
B. 179.2 cc
C. 44.8 cc
D. 22.4 cc

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the number of moles of urea.
The formula to calculate the number of moles is:

\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]

Given:
- Mass of urea = 0.120 g
- Molar mass of urea (NH₂CONH₂) = 60 g/mol

Calculating the number of moles:

\[ \text{Number of moles of urea} = \frac{0.120 \, \text{g}}{60 \, \text{g/mol}} = 0.002 \, \text{moles} \]

### Step 2: Determine the moles of gases produced.
From the reaction:

\[ \text{NH}_2\text{CONH}_2 + 2\text{HNO}_2 \rightarrow \text{CO}_2 + 2\text{N}_2 + 3\text{H}_2\text{O} \]

From the stoichiometry of the reaction:
- 1 mole of urea produces 1 mole of CO₂ and 2 moles of N₂.

Thus, for 0.002 moles of urea:
- Moles of CO₂ produced = 0.002 moles
- Moles of N₂ produced = 2 × 0.002 = 0.004 moles

### Step 3: Consider the effect of passing the gases through aqueous sodium hydroxide.
When the gases are passed through aqueous sodium hydroxide, CO₂ will react with NaOH and dissolve, leaving only N₂ gas.

Thus, the moles of gas remaining after passing through NaOH = moles of N₂ = 0.004 moles.

### Step 4: Calculate the volume of N₂ at STP.
At STP (Standard Temperature and Pressure), 1 mole of gas occupies 22.4 liters.

Calculating the volume of N₂:

\[ \text{Volume of N}_2 = \text{Number of moles of N}_2 \times 22.4 \, \text{L/mol} \]

\[ \text{Volume of N}_2 = 0.004 \, \text{moles} \times 22.4 \, \text{L/mol} = 0.0896 \, \text{L} \]

### Step 5: Convert the volume from liters to cubic centimeters (cc).
Since 1 L = 1000 cc:

\[ \text{Volume in cc} = 0.0896 \, \text{L} \times 1000 = 89.6 \, \text{cc} \]

### Final Answer:
The final volume of N₂ gas at STP is **89.6 cc**.

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