Question:
What is the maximum number of liters of HNO3(g) that can be made from 14.88 L of H2O(g) and 10.66 L of NO2(g) at STP?
3 NO2(g) + H2O(g) ==> 2 HNO3(g) + NO(g)
I have no idea how to do this. Please help if you can.
Thank you Dr. Bob!
To solve this problem, you can use the concept of stoichiometry and the given balanced chemical equation.
Step 1: Convert the volumes of gases to moles.
Since the gases are at STP, you can use the ideal gas law to convert the volumes to moles. At STP, one mole of any gas occupies 22.4 liters.
For water (H2O(g)), 14.88 L of water is equal to:
14.88 L × (1 mol/22.4 L) = 0.665 moles of H2O
For nitrogen dioxide (NO2(g)), 10.66 L of NO2 is equal to:
10.66 L × (1 mol/22.4 L) = 0.476 moles of NO2
Step 2: Use the stoichiometry of the balanced chemical equation to determine the limiting reactant.
The balanced chemical equation states that 3 moles of NO2 react with 1 mole of H2O to produce 2 moles of HNO3.
From the given data, we can see that the ratio of moles of NO2 to moles of H2O is approximately:
0.476 moles NO2 / 0.665 moles H2O ≈ 0.716
Comparing this ratio to the stoichiometric ratio of 3, we can see that the ratio is less than 3. This means that H2O is the limiting reactant in this case.
Step 3: Calculate the maximum number of moles of HNO3 that can be produced.
Since H2O is the limiting reactant, we use its moles to calculate the moles of HNO3 that can be produced.
From the stoichiometry of the balanced chemical equation, we know that 1 mole of H2O produces 2 moles of HNO3.
Therefore, the maximum moles of HNO3 that can be produced from the given reactants is:
0.665 moles H2O × (2 moles HNO3/1 mole H2O) = 1.33 moles HNO3
Step 4: Convert moles of HNO3 to liters of HNO3.
Finally, to convert moles of HNO3 to liters, you can use the ideal gas law again.
1 mole of any gas at STP occupies 22.4 liters of volume.
Therefore, the maximum liters of HNO3 that can be produced is:
1.33 moles HNO3 × (22.4 L/1 mole) = 29.7 L of HNO3
So, the maximum number of liters of HNO3(g) that can be made from 14.88 L of H2O(g) and 10.66 L of NO2(g) at STP is approximately 29.7 liters.
Here is a worked example of a simple stoichiometry problem. I work these limiting reagent problems (you know it is limiting reagent when BOTH reactants are given) by solving TWO of the simple ones.
http://www.jiskha.com/science/chemistry/stoichiometry.html
So convert L NO2 (moles = L/22.4) to moles and calculate moles HNO3 formed. Then convert L H2O to moles and calculate moles HNO3 formed. In limiting regent problems, the cxorrect answer (you have two answers) is ALWAYS the smaller value and the reagent producing that value is the limiting reagent. Then convert moles to L and you have it. BUT, there is an easier way.
When everything is in the gaseous form, you can work with L directly (you change L to moles in the first approach, then change moles back to liters---so we just shorten that and don't convert in the first place).
So 10.66 L NO2 x (2 moles HNO3/3 moles NO2) = 10.66 x (2/3) = 7.11 L HNO3.
Then 14.88 L H2O x (2 moles HNO3/1 mole H2O) = 14.88 x (2/1) = 29.7 L HNO3.
So the correct answer is you can form a maximum of 7.11 L HNO3.