How many grams of nitric acid, HNO3, can be prepared from the reaction of 207 g of NO2 with 54.0 g H2O according to the equation below?3NO2 + H2O → 2HNO3 + NOGroup of answer choices12692108189279

To solve this problem, we need to use stoichiometry, which is a method in chemistry that uses relationships between reactants and/or products in a chemical reaction to determine desired quantitative data.

Here are the steps:

1. First, we need to calculate the molar mass of NO2 and HNO3. The molar mass of NO2 is approximately 46.01 g/mol (14.01 g/mol for nitrogen and 16.00 g/mol for oxygen times 2). The molar mass of HNO3 is approximately 63.01 g/mol (1.01 g/mol for hydrogen, 14.01 g/mol for nitrogen, and 16.00 g/mol for oxygen times 3).

2. Next, we need to convert the given mass of NO2 and H2O to moles. We do this by dividing the given mass by the molar mass. For NO2, this is 207 g / 46.01 g/mol = 4.5 mol. For H2O, this is 54.0 g / 18.02 g/mol = 3.0 mol.

3. Now, we need to determine the limiting reactant, which is the reactant that will be completely consumed first in the reaction. According to the balanced chemical equation, 3 moles of NO2 react with 1 mole of H2O to produce 2 moles of HNO3. Therefore, if we have 4.5 moles of NO2, we would need 1.5 moles of H2O (4.5 mol / 3). However, we only have 3.0 moles of H2O, which is more than enough. Therefore, NO2 is the limiting reactant.

4. Finally, we can calculate the mass of HNO3 that can be produced. According to the balanced chemical equation, 3 moles of NO2 produce 2 moles of HNO3. Therefore, if we have 4.5 moles of NO2, we can produce 3.0 moles of HNO3 (4.5 mol * 2 / 3). To convert this to grams, we multiply by the molar mass of HNO3: 3.0 mol * 63.01 g/mol = 189 g.

So, 189 grams of nitric acid, HNO3, can be prepared from the reaction of 207 g of NO2 with 54.0 g H2O.