Nitric acid, , is manufactured by the Ostwald process, in which nitrogen dioxide, , reacts with water. How many grams of nitrogen dioxide are required in this reaction to produce 6.40 g ? g

7.04 g

Explanation:

Let's consider the reaction in the last step of the Ostwald process.

3 NO₂(g) + H₂O(l) → 2 HNO₃(aq) + NO(g)

The molar mass of HNO₃ is 63.01 g/mol. The moles corresponding to 6.40 g are:

6.40 g × (1 mol/63.01 g) = 0.102 mol

The molar ratio of NO₂ to HNO₃ is 3:2. The reacting moles of NO₂ are:

0.102 mol HNO₃ × (3 mol NO₂/2 mol HNO₃) = 0.153 mol NO₂

The molar mass of NO₂ is 46.01 g/mol. The mass corresponding to 0.153 moles is:

0.153 mol × (46.01 g/mol) = 7.04 g

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