Question

Nitric acid can be prepared by the reaction between NO2 and water producing nitric acid and NO. calculate the mass of NO2 gas required to produce 250mL of 1.5 molar nitric acid.

Answer 1

To produce 250 mL of 1.5 M nitric acid, approximately 25.88 grams of NO2 gas is required, based on the stoichiometry of the reaction between NO2 and water.

Step-by-step explanation:

In this chemistry question, we are calculating the mass of NO2 gas required to produce a specific volume and molarity of nitric acid. To find the mass of NO2 needed to produce 250 mL of 1.5 M nitric acid, we first need to calculate the number of moles of nitric acid required. The molarity (M) is defined as moles of solute per liter of solution, so for 250 mL (0.250 L) of 1.5 M solution, the moles of HNO3 needed are:

Number of moles = Molarity × Volume (in liters)

= 1.5 moles/L × 0.250 L

= 0.375 moles of HNO3

According to the balanced chemical reaction 3 NO2(g) + H2O(l) → 2 HNO3(aq) + NO(g), 3 moles of NO2 produce 2 moles of HNO3.

Therefore, to find the moles of NO2 required, we set up a proportion:

Moles of NO2 = (3 moles NO2 / 2 moles HNO3) × 0.375 moles HNO3

= 0.5625 moles of NO2

Now, we convert moles of NO2 to mass using the molar mass of NO2, which is approximately 46.0055 g/mol:

Mass of NO2 = Moles of NO2 × Molar mass of NO2

= 0.5625 moles × 46.0055 g/mol

≈ 25.88 g

Therefore, approximately 25.88 grams of NO2 gas is required to produce 250 mL of 1.5 M nitric acid.