To solve the problem step by step, we will follow the sequence of reactions and calculations as outlined in the video transcript.
### Step 1: Write the balanced chemical equation
When urea (NH2CO) is heated with sodium hydroxide (NaOH), it decomposes to form ammonia (NH3) and sodium carbonate (Na2CO3). The balanced chemical equation is:
\[ \text{2 NaOH} + \text{NH}_2\text{CO} \xrightarrow{\Delta} \text{Na}_2\text{CO}_3 + \text{2 NH}_3 \]
### Step 2: Calculate moles of urea
Given that we have 0.6 g of urea, we need to convert this mass into moles. The molar mass of urea (NH2CO) is approximately 60 g/mol.
\[
\text{Moles of urea} = \frac{\text{mass}}{\text{molar mass}} = \frac{0.6 \, \text{g}}{60 \, \text{g/mol}} = 0.01 \, \text{moles}
\]
### Step 3: Calculate moles of ammonia produced
From the balanced equation, we see that 1 mole of urea produces 2 moles of ammonia. Therefore, if we have 0.01 moles of urea, the moles of ammonia produced will be:
\[
\text{Moles of NH}_3 = 2 \times \text{moles of urea} = 2 \times 0.01 = 0.02 \, \text{moles}
\]
### Step 4: Determine the amount of HCl needed for neutralization
Ammonia (NH3) is a base and will react with hydrochloric acid (HCl) in a 1:1 mole ratio. Therefore, to completely neutralize 0.02 moles of NH3, we need 0.02 moles of HCl.
### Step 5: Analyze the options for HCl solution
We need to find which of the given HCl solutions contains exactly 0.02 moles of HCl. The formula to calculate moles from normality and volume is:
\[
\text{Number of moles} = \text{Normality} \times \text{Volume (L)}
\]
### Step 6: Calculate moles for each option
Assuming we have the following options (as an example):
1. 0.2 N HCl, 100 mL
2. 0.4 N HCl, 100 mL
3. 0.1 N HCl, 200 mL
4. 0.1 N HCl, 100 mL
**Option 1:**
\[
\text{Moles} = 0.2 \, \text{N} \times \frac{100 \, \text{mL}}{1000} = 0.2 \times 0.1 = 0.02 \, \text{moles}
\]
**Option 2:**
\[
\text{Moles} = 0.4 \, \text{N} \times \frac{100 \, \text{mL}}{1000} = 0.4 \times 0.1 = 0.04 \, \text{moles}
\]
**Option 3:**
\[
\text{Moles} = 0.1 \, \text{N} \times \frac{200 \, \text{mL}}{1000} = 0.1 \times 0.2 = 0.02 \, \text{moles}
\]
**Option 4:**
\[
\text{Moles} = 0.1 \, \text{N} \times \frac{100 \, \text{mL}}{1000} = 0.1 \times 0.1 = 0.01 \, \text{moles}
\]
### Step 7: Conclusion
From the calculations, options 1 and 3 provide the required 0.02 moles of HCl to completely neutralize the ammonia produced.
