Write a balanced equation and calculate the mass of nitrogen monoxide produced when 68.3 g of nitrogen dioxide reacts with excess water. (0.2 pts.)

Question

5. Nitrogen dioxide (MM = 46.01 g/mol) reacts with water (MM = 18.02 g/mol) to produce nitric acid (HNO3, MM = 63.02 g/mol) and nitrogen monoxide (MM = 30.01 g/mol). Write a balanced equation and calculate the mass of nitrogen monoxide produced when 68.3 g of nitrogen dioxide reacts with excess water. (0.2 pts.)

Answer 1

This is a stoichiometric problem, so we first need to write a balanced chemical equation for the reaction of nitrogen dioxide with water to form nitric acid and nitrogen monoxide. Nitrogen dioxide is an important chemical in the industrial production of nitric acid!

Here is the unbalanced reaction:

_NO₂ + _H₂O -> _HNO₃ + _NO

Let's start with the hydrogen. There are 2 hydrogens on the reactant side and 1 on the product side. Let's put a 2 for the stoichiometric number of nitric acids on the product side to balance the hydrogens:

_NO₂ + _H₂O -> 2HNO₃ + _NO

Next let's balance the nitrogens. There is 1 nitrogen on the reactant side and 3 on the product side. Let's put a 3 in front of nitrogen dioxide:

3NO₂ + _H₂O -> 2HNO₃ + _NO

Let's see where that leaves us: we have the same number of hydrogen (2) and nitrogen (3) on each side. There are (3 × 2) + 1 = 7 oxygens on the reactant side and (2 × 3) + 1 = 7 oxygens on the product side. We can now say that our equation is balanced!

3NO₂ + H₂O -> 2HNO₃ + NO

We can use this balanced equation to first calculate the number of moles of NO produced when 68.3 grams of NO₂ reacts with excess water. Let's begin by converting the mass of NO₂ into moles using the molar mass:

68.3 g NO₂ ÷ (46.01 g NO₂ / moles NO₂) = 1.48 moles NO₂ (3 significant figures).

Now we can use the mole ratio of NO₂ to NO in our balanced equation to calculate the number of moles of NO produced:

(2 moles NO ÷ 3 moles NO₂) × 1.48 moles NO₂ = 0.986 moles NO

The final step is to convert the moles of NO produced in the reaction to grams using the molar mass of NO:

0.986 moles NO × (30.01 g NO/ moles NO) = 29.6 g NO

68.3 grams of NO₂ reacts with excess water to produce 29.6 grams NO.