Nitric acid can be prepared from ammonia by the following steps;

4nh3 + 5o2 = 6h2o + 4no
2no + O2 = 2no2
2no2 + h2o = hno3 + hno2
3hno2 = hno3 + 2no + h2o
If all the steps take place consecutively, completely and without recycling, what mass of ammonia is needed to give 6.3g of nitric acid.

To solve this problem, we need to calculate the mass of ammonia required to produce 6.3g of nitric acid. Let's break down the problem into the following steps:

Step 1: Calculate the molar mass of nitric acid (HNO3):
The molar mass of HNO3 can be calculated by adding up the atomic masses of hydrogen (H), nitrogen (N), and oxygen (O) in one mole of nitric acid.
Molar mass of HNO3:
= (1 * atomic mass of H) + (1 * atomic mass of N) + (3 * atomic mass of O)
= (1 * 1.01 g/mol) + (1 * 14.01 g/mol) + (3 * 16.00 g/mol)
= 1.01 g/mol + 14.01 g/mol + 48.00 g/mol
= 63.02 g/mol

Step 2: Calculate the moles of HNO3:
The number of moles of HNO3 can be calculated using the given mass of nitric acid and the molar mass of HNO3.
Moles of HNO3 = Mass of HNO3 / Molar mass of HNO3
= 6.3 g / 63.02 g/mol
= 0.1 mol

Step 3: Determine the stoichiometry between ammonia (NH3) and nitric acid (HNO3):
From the balanced chemical equations provided, we can see that 4 moles of NH3 produce 3 moles of HNO3.

Step 4: Calculate the moles of NH3 required:
Using the stoichiometry ratio between NH3 and HNO3, we can calculate the moles of NH3 needed to produce the given moles of HNO3.
Moles of NH3 = (Moles of HNO3 * 4) / 3
= (0.1 * 4) / 3
≈ 0.133 mol

Step 5: Convert moles of NH3 to grams:
Finally, we can calculate the mass of ammonia required by multiplying the moles of NH3 by its molar mass.
Mass of NH3 = Moles of NH3 * Molar mass of NH3
= 0.133 mol * (1 * 14.01 g/mol)
≈ 1.86 g

Therefore, approximately 1.86 grams of ammonia is needed to produce 6.3 grams of nitric acid.