Question

Given the following equation:

3 NO₂ (g)+ H₂O(l)2HNO₃ (aq)+ NO(g)
Assume STP
a. What mass of water is required to react with 15.5 L of Nitrogen dioxide?
b. What volume of Nitrogen monoxide would be produced from 100.0 g of water?
c. If 42.0 L of NO(g)
is produced, what volume of NO₂ (g)
reacted?
d. How many water atoms would be required to react with an excess of
Nitrogen dioxide to produce 46.0 L of Nitrogen monoxide?

Answer 1

Chemical reaction stoichiometry allows calculation of masses and volumes of reactants and products based on molar ratios. Using the balanced reaction and the ideal gas law, quantities related to water and nitrogen compounds can be determined through conversion between moles, mass, and volume.

Step-by-step explanation:

Stoichiometry Calculations Based on Chemical Reactions

To answer this student's questions, stoichiometry as well as the ideal gas law and molar relationships between reactants and products are applied. Here are the solutions:

• a. The mass of water required to react with 15.5 L of Nitrogen dioxide (Given the balanced equation, the molar ratio is 3:1 NO2 to H2O. Using the molar volume at STP, 22.4 L/mol, we can calculate the moles of NO2 and from there determine the required mass of H2O).
• b. The volume of Nitrogen monoxide produced from 100.0 g of water (This can be determined using the molar ratios from the balanced equation and converting from mass of water to moles, and finally to the volume of NO using the ideal gas law).
• c. If 42.0 L of NO(g) is produced, the volume of NO2(g) reacted can be calculated by using the molar ratio from the balanced equation and the molar volume at STP).
• d. The number of water molecules required to react with an excess of Nitrogen dioxide to produce 46.0 L of Nitrogen monoxide (First calculate the moles of NO given the volume at STP, then use the balanced equation to find the moles of H2O required, and finally multiply by Avogadro's number to find the number of molecules).

Answer 2

a. 92.08 g H2O
b. 124.353 L NO
c. 125.647 L NO2
d. 1.233 x 10^24 atoms H2O

Step-by-step explanation:

To answer each part of the question:

a. To determine the mass of water required to react with 15.5 L of nitrogen dioxide (NO2), we need to first balance the equation:
3 NO2 (g) + H2O (l) -> 2 HNO3 (aq) + NO (g)

From the balanced equation, we can see that the molar ratio of NO2 to H2O is 3:1. Therefore, for every 3 moles of NO2, 1 mole of H2O is needed. To find the mass of water, we can use the molar mass of water (18.015 g/mol) and the molar ratio:
15.5 L NO2 * (1 mol NO2 / 22.414 L) * (1 mol H2O / 3 mol NO2) * (18.015 g H2O / 1 mol H2O) = 92.08 g H2O

b. To determine the volume of nitrogen monoxide (NO) produced from 100.0 g of water, we need to first determine the moles of water:
100.0 g H2O / 18.015 g/mol = 5.55 mol H2O

From the balanced equation, we can see that the molar ratio of H2O to NO is 1:1. Therefore, for every 1 mole of H2O, 1 mole of NO is produced. To find the volume of NO, we can use the molar volume of a gas at STP (22.414 L/mol) and the molar ratio:
5.55 mol H2O * (1 mol NO / 1 mol H2O) * (22.414 L NO / 1 mol NO) = 124.353 L NO

c. To determine the volume of NO2 that reacted to produce 42.0 L of NO, we need to first determine the moles of NO:
42.0 L NO * (1 mol NO / 22.414 L) = 1.875 mol NO

From the balanced equation, we can see that the molar ratio of NO to NO2 is 1:3. Therefore, for every 1 mole of NO, 3 moles of NO2 react. To find the volume of NO2, we can use the molar volume of a gas at STP (22.414 L/mol) and the molar ratio:
1.875 mol NO * (3 mol NO2 / 1 mol NO) * (22.414 L NO2 / 1 mol NO2) = 125.647 L NO2

d. To determine the number of water atoms required to react with an excess of nitrogen dioxide to produce 46.0 L of nitrogen monoxide, we need to first determine the moles of NO:
46.0 L NO * (1 mol NO / 22.414 L) = 2.054 mol NO

From the balanced equation, we can see that the molar ratio of NO to H2O is 1:1. Therefore, for every 1 mole of NO, 1 mole of H2O is required. To find the number of water atoms, we can use Avogadro's number (6.022 x 10^23 atoms/mol) and the molar ratio:
2.054 mol NO * (1 mol H2O / 1 mol NO) * (6.022 x 10^23 atoms H2O / 1 mol H2O) = 1.233 x 10^24 atoms H2O