At constant temperature and pressure, what is the maximum volume, in liters, of HNO3(g) that can be made from 10.39 L of NO2(g) and 15.53 L of H2O(g)? I don't even know how to start it...
NO2 is not the anhydride of HNO3. How advanced is this course? Do you know what equation you are expected to use?
3 NO2(g) + H2O(g) 2 HNO3(g) + NO(g)
At constant temperature and pressure, what is the maximum volume, in liters, of HNO3(g) that can be made from 10.39 L of NO2(g) and 15.53 L of H2O(g)?
That is the entire question and this is AP Chem but gas stoichiomatry from regular chem. I just do not remember at all how to do this
Thanks for the added information. This is a limiting reagent problem. All of them can be worked the same way. Here is a worked example. Let me know if you get stuck but post your work if you do.
http://www.jiskha.com/science/chemistry/limiting_reagent_problem.html
To solve this problem, you need to use the balanced chemical equation for the reaction between NO2(g) and H2O(g) to form HNO3(g). The equation for this reaction is:
3 NO2(g) + H2O(g) → 2 HNO3(g)
To find the maximum volume of HNO3(g) that can be produced, you need to determine which reactant is limiting and calculate the volume of HNO3(g) based on that limiting reactant.
Here's how you can approach the problem step by step:
1. Convert the given volumes of NO2(g) and H2O(g) into moles using the ideal gas law:
Moles of NO2(g) = (10.39 L) / (22.4 L/mol) (since 1 mole occupies 22.4 L at STP)
Moles of H2O(g) = (15.53 L) / (22.4 L/mol)
2. Determine the limiting reactant by comparing the number of moles of NO2(g) and H2O(g) using the stoichiometry of the balanced equation. The reactant that produces fewer moles of HNO3(g) will be the limiting reactant.
In this case, you need to find the ratio of moles of NO2(g) to moles of HNO3(g) and the ratio of moles of H2O(g) to moles of HNO3(g) using the balanced equation:
NO2(g):HNO3(g) = 3:2
H2O(g):HNO3(g) = 1:2
Calculate the moles of HNO3(g) that can be produced by each reactant using the ratios:
Moles of HNO3(g) from NO2(g) = (Moles of NO2(g)) * (2 mol HNO3(g) / 3 mol NO2(g))
Moles of HNO3(g) from H2O(g) = (Moles of H2O(g)) * (2 mol HNO3(g) / 1 mol H2O(g))
3. Compare the moles of HNO3(g) produced by each reactant. The smaller value will be the limiting reactant.
4. Once you have identified the limiting reactant, use the stoichiometry of the balanced equation to calculate the moles of HNO3(g) that can be produced.
5. Finally, convert the moles of HNO3(g) to volume using the ideal gas law:
Volume of HNO3(g) = (Moles of HNO3(g)) * (22.4 L/mol) (inverse of the first step)
This approach will help you find the maximum volume of HNO3(g) that can be produced from the given volumes of NO2(g) and H2O(g) at constant temperature and pressure.