3 NO2(g) + H2O(g) ==> 2 HNO3(g) + NO(g)
What is the maximum number of liters of HNO3(g) that can be made from 17.45 L of H2O(g) and 12.06 L of NO2(g) at STP?
To find the maximum number of liters of HNO3(g) that can be made from the given reactants at STP, we need to determine the limiting reactant first. The limiting reactant is the reactant that is completely consumed in the reaction and determines the maximum amount of product that can be formed.
To find the limiting reactant, we compare the moles of each reactant to the stoichiometry of the balanced equation:
From the balanced equation, we see that the ratio of NO2 to HNO3 is 3:2. Assuming STP conditions, 1 mole of any gas occupies 22.4 liters. Therefore, we can convert the given volumes to moles:
- Moles of H2O(g) = 17.45 L / 22.4 L/mol = 0.7790 mol
- Moles of NO2(g) = 12.06 L / 22.4 L/mol = 0.5384 mol
Now, we compare the number of moles of each reactant to the stoichiometric ratio in the balanced equation:
- For H2O(g): 0.7790 mol H2O x (2 mol HNO3 / 1 mol H2O) = 1.558 mol HNO3
- For NO2(g): 0.5384 mol NO2 x (2 mol HNO3 / 3 mol NO2) = 0.3589 mol HNO3
From the calculations, we can see that the moles of HNO3 produced from H2O is greater than from NO2. Therefore, NO2 is the limiting reactant.
To find the maximum number of liters of HNO3, we can use the stoichiometry of the balanced equation:
From the equation, we see that 3 moles of NO2 produce 2 moles of HNO3. Therefore, we can calculate the moles of HNO3 produced from NO2:
- Moles of HNO3 = 0.3589 mol NO2 x (2 mol HNO3 / 3 mol NO2) = 0.2392 mol HNO3
Now we can convert the moles of HNO3 to liters:
- Liters of HNO3 = 0.2392 mol HNO3 x (22.4 L/mol) = 5.35 L HNO3
Therefore, the maximum number of liters of HNO3 that can be made from 17.45 L of H2O(g) and 12.06 L of NO2(g) at STP is 5.35 L HNO3.