`N_(2)H_(4)`, Hydrazine a rocket fuel can be produced according to the following reaction :
`ClNH_(2) + 2NH_(3) rarrN_(2)H_(4) + NH_(4)CI`
When `1000 g CINH_(2)` is reacted with excess of `NH_(3),473 g N_(2)H_(4)` is produced. What is the `%` yield of the reaction.

Explanation

To calculate the percentage yield of the reaction producing hydrazine (N₂H₄) from chloramine (ClNH₂) and ammonia (NH₃), we will follow these steps:

### Step 1: Calculate the molar mass of ClNH₂
- Chlorine (Cl) = 35.5 g/mol
- Nitrogen (N) = 14 g/mol
- Hydrogen (H) = 1 g/mol (2 atoms)

Molar mass of ClNH₂ = 35.5 + 14 + (1 × 2) = 35.5 + 14 + 2 = 51.5 g/mol

### Step 2: Calculate the molar mass of N₂H₄
- Nitrogen (N) = 14 g/mol (2 atoms)
- Hydrogen (H) = 1 g/mol (4 atoms)

Molar mass of N₂H₄ = (14 × 2) + (1 × 4) = 28 + 4 = 32 g/mol

### Step 3: Calculate the number of moles of ClNH₂
Given mass of ClNH₂ = 1000 g

Number of moles of ClNH₂ = mass / molar mass
= 1000 g / 51.5 g/mol
≈ 19.417 moles

### Step 4: Calculate the number of moles of N₂H₄ produced
Given mass of N₂H₄ produced = 473 g

Number of moles of N₂H₄ = mass / molar mass
= 473 g / 32 g/mol
≈ 14.781 moles

### Step 5: Determine the theoretical yield of N₂H₄
From the balanced equation, 1 mole of ClNH₂ produces 1 mole of N₂H₄. Therefore, the theoretical yield of N₂H₄ from 19.417 moles of ClNH₂ is also 19.417 moles.

### Step 6: Calculate the percentage yield
Percentage yield = (actual yield / theoretical yield) × 100%
= (14.781 moles / 19.417 moles) × 100%
≈ 76.13%

### Final Answer:
The percentage yield of the reaction is approximately **76.13%**.

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