The ostwald process is used commercially to produce nitric acid, which is, in turn, used in many modern chemical processes. in the first step of the ostwald process, ammonia is reacted with oxygen gas to produce nitric oxide and water. what is the maximum mass of h2oh2o that can be produced by combining 70.170.1 g of each reactant?

Explanation:

We are given the mass of two reactants, so this is a limiting reactant problem.

We know that we will need mases, moles, and molar masses, so, let's assemble all the data in one place, with molar masses above the formulas and masses below them.

M_r:    17.03   32.00                 18.02  

           4NH₃ + 5O₂ ⟶ 4NO + 6H₂O

m/g:     70.1      70.1

Step 1. Calculate the moles of each reactant

Step 2. Identify the limiting reactant  

Calculate the moles of H₂O we can obtain from each reactant.

From NH₃:

The molar ratio of H₂O:NH₃ is 6:4.

From O₂:  

The molar ratio of H₂O:O₂ is 6:5.  

O₂ is the limiting reactant because it gives the smaller amount of H₂O.  

Step 3. Calculate the theoretical yield.

in scientific notation, the value of the  speed of light in a vacuum

  will be written as 2.995 * 10^8. when this value is represented in the form of

  a * 10^b, then, a = 2.995 and b = 8.

the value of the speed of light in a vacuum given above can also be rounded up to approximately 300000000. in this case, the scientific notation will be written as 3 *10^8.