How many grams of nitric acid can be prepared from 50.0 g of `KNO_3` of 80% purity ?

Explanation

To determine how many grams of nitric acid (HNO3) can be prepared from 50.0 g of potassium nitrate (KNO3) with 80% purity, we can follow these steps:

### Step 1: Calculate the actual mass of KNO3
Given that the sample has 80% purity, we need to find the actual mass of KNO3 in the 50.0 g sample.

\[ \text{Actual mass of KNO3} = \text{Total mass} \times \text{Purity} \]
\[ \text{Actual mass of KNO3} = 50.0 \, \text{g} \times 0.80 = 40.0 \, \text{g} \]

### Step 2: Calculate the number of moles of KNO3
Next, we need to find the number of moles of KNO3. The molar mass of KNO3 is approximately 101 g/mol.

\[ \text{Moles of KNO3} = \frac{\text{Mass of KNO3}}{\text{Molar mass of KNO3}} \]
\[ \text{Moles of KNO3} = \frac{40.0 \, \text{g}}{101 \, \text{g/mol}} \approx 0.396 \, \text{mol} \]

### Step 3: Use stoichiometry to find moles of HNO3 produced
From the balanced chemical reaction, we know that 1 mole of KNO3 produces 1 mole of HNO3. Therefore, the moles of HNO3 produced will be the same as the moles of KNO3.

\[ \text{Moles of HNO3} = \text{Moles of KNO3} = 0.396 \, \text{mol} \]

### Step 4: Calculate the mass of HNO3 produced
Now we can find the mass of HNO3 produced using its molar mass, which is approximately 63 g/mol.

\[ \text{Mass of HNO3} = \text{Moles of HNO3} \times \text{Molar mass of HNO3} \]
\[ \text{Mass of HNO3} = 0.396 \, \text{mol} \times 63 \, \text{g/mol} \approx 24.948 \, \text{g} \]

### Step 5: Round the result
Finally, we can round the result to a suitable number of significant figures.

\[ \text{Mass of HNO3} \approx 25.0 \, \text{g} \]

### Conclusion
From 50.0 g of KNO3 with 80% purity, approximately 25.0 g of nitric acid can be prepared.

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