How many grams of urea on heating yield `10^(22)` molecules of biuret by the reaction :
`2CO(NH_(2))_(2) rarr H_(2)N - CO - NH - CO - NH_(2) + NH_(3)` ?

Options

A. `1.495`
B. `0.995`
C. `1.99`
D. `1.753`

Explanation

To solve the problem of how many grams of urea yield \(10^{22}\) molecules of biuret upon heating, we can follow these steps:

### Step 1: Convert molecules of biuret to moles
To find the number of moles of biuret, we use Avogadro's number, which is approximately \(6.02 \times 10^{23}\) molecules/mol.

\[ \text{Moles of biuret} = \frac{10^{22} \text{ molecules}}{6.02 \times 10^{23} \text{ molecules/mol}} \]

Calculating this gives:

\[ \text{Moles of biuret} \approx 0.0166 \text{ moles} \]

### Step 2: Relate moles of biuret to moles of urea
From the balanced chemical equation:

\[ 2 \text{CO}(NH_2)_2 \rightarrow \text{H}_2N - CO - NH - CO - NH_2 + NH_3 \]

We see that 2 moles of urea produce 1 mole of biuret. Therefore, to find the moles of urea needed, we multiply the moles of biuret by 2:

\[ \text{Moles of urea} = 2 \times \text{Moles of biuret} = 2 \times 0.0166 \approx 0.0332 \text{ moles} \]

### Step 3: Calculate the mass of urea
The molar mass of urea \((CO(NH_2)_2)\) is calculated as follows:

- Carbon (C): 12 g/mol
- Oxygen (O): 16 g/mol
- Nitrogen (N): 14 g/mol (2 Nitrogens)
- Hydrogen (H): 1 g/mol (4 Hydrogens)

Calculating the molar mass:

\[ \text{Molar mass of urea} = 12 + 16 + (2 \times 14) + (4 \times 1) = 12 + 16 + 28 + 4 = 60 \text{ g/mol} \]

Now, we can find the mass of urea needed:

\[ \text{Mass of urea} = \text{Moles of urea} \times \text{Molar mass of urea} = 0.0332 \text{ moles} \times 60 \text{ g/mol} \]

Calculating this gives:

\[ \text{Mass of urea} \approx 1.992 \text{ grams} \]

### Final Answer
Thus, the mass of urea required to yield \(10^{22}\) molecules of biuret is approximately **1.99 grams**.

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