Question

What is the mass in grams of NO that will be produced from 19.5 g of NO₂​ reacted with excess water in the following chemical reaction? 3NO₂(g)+H₂​O(l)→2HNO₃(g)+NO(g) A) 7.5 g B) 9.0 g C) 13.0 g D) 15.0 g

Answer 1

The mass of NO produced from 19.5 g of NO2, using stoichiometry and the balanced chemical reaction with excess water, is calculated to be approximately 4.24 grams. This result is not reflected in the given multiple choices.

Step-by-step explanation:

Calculating the Mass of NO Produced

To calculate the mass of NO (nitric oxide) that will form from the reaction of 19.5 grams of NO2 with excess water, we will use stoichiometry and the given balanced chemical equation:

3NO2(g) + H2O(l) → 2HNO3(g) + NO(g)

1. Determine the molar mass of NO2, which is 14.01 (N) + 32.00 (O*2) = 46.01 g/mol.
2. Calculate the number of moles of NO2 using the mass of NO2 provided: 19.5 g NO2 / 46.01 g/mol = 0.4239 moles of NO2.
3. Use the mole ratio from the balanced equation to find the moles of NO. Since the ratio is 3 moles of NO2 to 1 mole of NO, this would yield 0.4239 moles NO2 * (1 mole NO / 3 moles NO2) =
   0.1413 moles of NO.
4. Determine the molar mass of NO, which is 14.01 (N) + 16.00 (O) = 30.01 g/mol.
5. Calculate the mass of NO produced using its molar mass: 0.1413 moles NO * 30.01 g/mol = 4.24 grams of NO.

Based on these calculations, the mass of NO produced from 19.5 g of NO2 is approximately 4.24 grams, which is not an option in the multiple choices provided, possibly due to a typo in the question. It's important to use the correct molar masses and stoichiometry to find an accurate result based on a balanced chemical equation.