# How can I calculate the volume of water produced during this reaction: HNO3 + KOH -> H2O + KNO3 The reaction between nitric acid and potassium hydroxide is:

`HNO_3 + KOH -> H_2O + KNO_3`

In this reaction, nitric acid and potassium hydroxide react to form water and potassium nitrate.

Since the equation is balanced, we can use stoichiometry to calculate the quantities of various chemicals required or formed.

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The reaction between nitric acid and potassium hydroxide is:

`HNO_3 + KOH -> H_2O + KNO_3`

In this reaction, nitric acid and potassium hydroxide react to form water and potassium nitrate.

Since the equation is balanced, we can use stoichiometry to calculate the quantities of various chemicals required or formed.

From the equation, 1 mole of nitric acid reacts with 1 mole of potassium hydroxide to generate 1 mole of water and 1 mole of potassium nitrate. Thus, given the amount of any of the chemicals, we can determine the amount of water formed.

For example, say 63 g of nitric acid (molar mass = 63 g/mole) is mixed with excess potassium hydroxide, then 1 mole of nitric acid (63 g / 63 g/mole) will produce 1 mole of water (molar mass = 18 g/mole) or 18 g of water. Since the water has a density of about 1 g/ml, 18 g or 18 ml of water will be produced.

Hope this helps.

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