A solid mixture(5.000 g) consisting of lead nitrate and sodium nitrate was heated below 600°C until the weight of the residue was constant. If the loss in weight is 28%, find the amount of lead nitrate and sodium nitrate in the mixture.

Explanation

To solve the problem of finding the amounts of lead nitrate (Pb(NO3)2) and sodium nitrate (NaNO3) in a 5.000 g mixture that loses 28% of its weight upon heating, we can follow these steps:

### Step 1: Define Variables
Let:
- \( x \) = mass of lead nitrate (Pb(NO3)2) in grams
- \( y \) = mass of sodium nitrate (NaNO3) in grams

### Step 2: Set Up the Mass Equation
Since the total mass of the mixture is 5.000 g, we can write the equation:
\[ x + y = 5 \quad \text{(Equation 1)} \]

### Step 3: Calculate Weight Loss from Lead Nitrate
The molar mass of lead nitrate (Pb(NO3)2) is 331 g/mol. When heated, it decomposes as follows:
\[ \text{Pb(NO3)2} \rightarrow \text{PbO} + 2 \text{NO2} + \frac{1}{2} \text{O2} \]
The total mass loss from this reaction is:
- Loss from 2 moles of NO2: \( 2 \times 46 \, \text{g} = 92 \, \text{g} \)
- Loss from \( \frac{1}{2} \) mole of O2: \( \frac{1}{2} \times 32 \, \text{g} = 16 \, \text{g} \)
- Total loss = \( 92 + 16 = 108 \, \text{g} \)

Thus, for \( x \) grams of lead nitrate, the weight loss is:
\[ \text{Weight loss from Pb(NO3)2} = \frac{108}{331} \times x \]

### Step 4: Calculate Weight Loss from Sodium Nitrate
The molar mass of sodium nitrate (NaNO3) is 85 g/mol. When heated, it decomposes as follows:
\[ \text{NaNO3} \rightarrow \text{NaNO2} + \frac{1}{2} \text{O2} \]
The total mass loss from this reaction is:
- Loss from \( \frac{1}{2} \) mole of O2: \( \frac{1}{2} \times 32 \, \text{g} = 16 \, \text{g} \)

Thus, for \( y \) grams of sodium nitrate, the weight loss is:
\[ \text{Weight loss from NaNO3} = \frac{16}{85} \times y \]

### Step 5: Set Up the Total Weight Loss Equation
The total weight loss from the mixture is given as 28% of 5 g:
\[ \text{Total weight loss} = 0.28 \times 5 = 1.4 \, \text{g} \]

Now, we can write the equation for total weight loss:
\[ \frac{108}{331} x + \frac{16}{85} y = 1.4 \quad \text{(Equation 2)} \]

### Step 6: Substitute Equation 1 into Equation 2
From Equation 1, we can express \( y \) in terms of \( x \):
\[ y = 5 - x \]

Substituting this into Equation 2 gives:
\[ \frac{108}{331} x + \frac{16}{85} (5 - x) = 1.4 \]

### Step 7: Solve for \( x \)
Expanding and simplifying:
\[ \frac{108}{331} x + \frac{80}{85} - \frac{16}{85} x = 1.4 \]
Converting \( \frac{80}{85} \) to a decimal gives approximately 0.9412.

Combining like terms:
\[ \left( \frac{108}{331} - \frac{16}{85} \right)x + 0.9412 = 1.4 \]
This simplifies to:
\[ \left( \frac{108}{331} - \frac{16}{85} \right)x = 1.4 - 0.9412 \]
Calculating the right-hand side gives approximately 0.4588.

Now, solving for \( x \):
\[ x \approx 3.33 \, \text{g} \]

### Step 8: Calculate \( y \)
Using Equation 1:
\[ y = 5 - x = 5 - 3.33 = 1.67 \, \text{g} \]

### Final Answer
- Mass of lead nitrate (Pb(NO3)2) = 3.33 g
- Mass of sodium nitrate (NaNO3) = 1.67 g