Question

How much Potassium Nitrate do I need to make 21g of Nitric Acid?

Answer 1

It takes 0.333 mol (33.7 g) of KNO₃ to make 21 g of HNO₃.

Step-by-step explanation

KNO₃ reacts with concentrated H₂SO₄ to produce HNO₃ and KHSO₄.

`\text{KNO}_3 + \text{H}_2\text{SO}_4 \; (\text{conc.}) \to \text{HNO}_3 + \text{KHSO}_4` (balanced.)

Each mole formula unit of KNO₃ reacts with excess H₂SO₄ to produce one mole of HNO₃. In other words, it takes the same number formula units of KNO₃ to produce a certain number of moles of HNO₃.

How many moles of molecules in 21 g of HNO₃?

Relative atomic mass:

• H: 1.008;
• N: 14.007;
• O: 15.999.

Molar mass of HNO₃:
`1.008 + 14.007 + 3 * 15.999 = 63.01 \; \text{g}\cdot \text{mol}^(-1)`.

`n(\text{HNO}_3) = \frac{m(\text{HNO}_3)}{M(\text{HNO}_3)} = \frac{21\; \text{g}}{63.012\; \text{g}\cdot \text{mol}^(-1)} = 0.333 \; \text{mol}`.

It takes 0.333 moles formula units of KNO₃ to produce 21 g or 0.333 moles of HNO₃.

What's the mass of 0.333 moles formula units of KNO₃?

Molar mass of KNO₃:
`39.098 + 14.007 + 3 * 15.999 = 101.10 \; \text{g}\cdot \text{mol}^(-1)`

`m(\text{KNO}_3) = n(\text{KNO}_3) \cdot M(\text{KNO}_3)\\\phantom{m(\text{KNO}_3)} = 0.333 \; \text{mol} * 101.10 \; \text{g}\cdot \text{mol}^(-1)\\\phantom{m(\text{KNO}_3)} = 33.7 \; \text{g}`.