Final answer:
To make 1100.0 L of a 0.200 M aqueous solution of nitric acid, assuming 100% yield, one must start with 3746.6 grams of ammonia (NH3).
Step-by-step explanation:
The reaction to form nitric acid (HNO3) from ammonia (NH3) involves several steps, starting with the oxidation of ammonia. The equation for the initial reaction is:
4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(l)
To find out how many grams of NH3 are required to produce 1100.0 L of a 0.200 M HNO3 solution, we first need to calculate the moles of HNO3 needed using the molarity formula:
Moles of HNO3 = Molarity × Volume = 0.200 mol/L × 1100.0 L = 220 moles
Since the stoichiometry of NH3 to HNO3 is 1:1 in the overall process (after considering all steps), you would also require 220 moles of NH3.
Now, we can convert moles of NH3 to grams using its molar mass (17.03 g/mol):
Mass of NH3 = 220 moles × 17.03 g/mol = 3746.6 grams
Therefore, you must start with 3746.6 grams of NH3 to make 1100.0 L of a 0.200 M aqueous solution of nitric acid, assuming 100% yield.