31.3 gram mixture NaBr and NaCl treated with `H_(2)SO_(4).28.4` gram of `Na_(2)SO_(4)` is produced. Then calculate the amount of NaCl and NaBr in the mixture.

Answer

To solve the problem of determining the amounts of NaCl and NaBr in a 31.3 gram mixture that produces 28.4 grams of Na2SO4 when treated with H2SO4, we can follow these steps:

### Step 1: Calculate the moles of Na2SO4 produced
First, we need to find the molar mass of Na2SO4:
- Sodium (Na) = 23 g/mol (2 Na = 46 g)
- Sulfur (S) = 32 g/mol
- Oxygen (O) = 16 g/mol (4 O = 64 g)

Adding these together:
\[ \text{Molar mass of Na2SO4} = 46 + 32 + 64 = 142 \text{ g/mol} \]

Now, we can calculate the moles of Na2SO4 produced:
\[ \text{Moles of Na2SO4} = \frac{\text{mass}}{\text{molar mass}} = \frac{28.4 \text{ g}}{142 \text{ g/mol}} = 0.2 \text{ moles} \]

### Step 2: Determine the moles of sodium (Na) in Na2SO4
Since Na2SO4 contains 2 moles of Na for every mole of Na2SO4:
\[ \text{Moles of Na} = 2 \times \text{moles of Na2SO4} = 2 \times 0.2 = 0.4 \text{ moles} \]

### Step 3: Set up the equations for the mixture
Let:
- \( x \) = mass of NaCl in grams
- \( 31.3 - x \) = mass of NaBr in grams

Next, we calculate the moles of NaCl and NaBr:
- Molar mass of NaCl = 58.5 g/mol
- Molar mass of NaBr = 103 g/mol

Moles of NaCl:
\[ \text{Moles of NaCl} = \frac{x}{58.5} \]

Moles of NaBr:
\[ \text{Moles of NaBr} = \frac{31.3 - x}{103} \]

### Step 4: Set up the equation for total moles of Na
The total moles of Na from both salts must equal the moles of Na produced:
\[ \frac{x}{58.5} + \frac{31.3 - x}{103} = 0.4 \]

### Step 5: Solve the equation
To solve for \( x \), we can multiply through by the least common multiple of the denominators (58.5 and 103) to eliminate the fractions:
\[ 103x + 58.5(31.3 - x) = 0.4 \times 58.5 \times 103 \]

Calculating the right side:
\[ 0.4 \times 58.5 \times 103 = 2407.8 \]

Expanding the left side:
\[ 103x + 1825.05 - 58.5x = 2407.8 \]

Combining like terms:
\[ 44.5x + 1825.05 = 2407.8 \]

Isolating \( x \):
\[ 44.5x = 2407.8 - 1825.05 \]
\[ 44.5x = 582.75 \]
\[ x = \frac{582.75}{44.5} \approx 13.1 \text{ g} \]

### Step 6: Calculate the mass of NaBr
Now, we can find the mass of NaBr:
\[ \text{Mass of NaBr} = 31.3 - x = 31.3 - 13.1 = 18.2 \text{ g} \]

### Final Answer
- Mass of NaCl = 13.1 g
- Mass of NaBr = 18.2 g

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