4NH3 + 5O2 → 4NO + 6H2O

2NO + O2 → 2NO2
3NO2 + H2O → 2HNO3 + NO

In this process, the NO made in the last reaction is recycled back into the second reaction and used as a reactant. In the end, all of the NO produced is recycled back so that when the manufacturing process is complete, essentially no NO remains. If a chemist starts with 50.0 grams of NH3 and an excess of water and oxygen, what mass of nitric acid can be made

Here is a worked example of a single stoichiometry problem. Just follow the steps. The process is the same for a multiple step problem.

http://www.jiskha.com/science/chemistry/stoichiometry.html

Consider the following equation, which represents the combustion of ammonia.

4NH3(g) + 3O2(g) --> 2N2(g) + 6H2O(g)

How many moles of O2(g) is required to form 10 mol of H2O(g) is burned?

To determine the mass of nitric acid (HNO3) that can be made, we need to find the limiting reactant in each step and calculate the amount of products formed.

Step 1:
We start with 50.0 grams of NH3. To find the amount of NH3 in moles, we need to divide the given mass by its molar mass. The molar mass of NH3 (ammonia) is 17.03 g/mol.
50.0 g NH3 / 17.03 g/mol = 2.93 mol NH3

From the balanced chemical equation, we can see that the ratio of NH3 to NO is 4:4. This means that for every 4 moles of NH3, we can produce 4 moles of NO.

Since we have 2.93 moles of NH3, we can calculate the number of moles of NO produced:
2.93 mol NH3 × (4 mol NO / 4 mol NH3) = 2.93 mol NO

Step 2:
In the second reaction, the NO produced in the previous step is recycled back and used as a reactant.

Since the ratio of NO to NO2 is 2:1, this means that for every 2 moles of NO, we can produce 1 mole of NO2.

Since we have 2.93 moles of NO produced in step 1, we can calculate the number of moles of NO2 produced:
2.93 mol NO × (1 mol NO2 / 2 mol NO) = 1.47 mol NO2

Step 3:
In the final reaction, 3 moles of NO2 react with water (H2O) to produce 2 moles of nitric acid (HNO3) and 1 mole of NO.

Since the ratio of NO2 to HNO3 is 3:2, this means that for every 3 moles of NO2, we can produce 2 moles of HNO3.

Since we have 1.47 moles of NO2 produced in step 2, we can calculate the number of moles of HNO3 produced:
1.47 mol NO2 × (2 mol HNO3 / 3 mol NO2) = 0.98 mol HNO3

To find the mass of HNO3 produced, we need to multiply the number of moles of HNO3 by its molar mass. The molar mass of HNO3 (nitric acid) is 63.02 g/mol.
0.98 mol HNO3 × 63.02 g/mol = 61.66 g HNO3

Therefore, the mass of nitric acid that can be made from 50.0 grams of NH3 is approximately 61.66 grams.