What mass of nh3 in grams must be used to produce 1.81 tons of hno3 by the ostwald process, assuming an 80.0 percent yield in each step (1 ton=2000 lb; 1 lb= 453.6 g)?

Mass of used is 1730812.5 grams.

Explanation: Ostwald's process is the process of the production of nitric acid from ammonia. Steps for the process are:

Total mass of produced = 1.81 tons

Using conversion factor, we get

Total mass of produced = (1.81 × 2000 × 453.6)g = 1642032 g

Moles of is calculated by using the formula:

....(1)

Molar mass of nitric acid = 63 g/mol

We are given that the percentage yield is 80 % in every step, then

Moles of , theorectically (From Step-III) = 2 × 26064 = 52128 moles

Actual moles of

Now, theoretical moles of NO (in Step-II) = Actual moles of

Theoretical moles of NO = 65160 moles

Actual moles of NO =

Now, theoretical moles of (in Step-I) = Actual moles of NO

Theoretical moles of = 81450 moles

Actual moles of =

To calculate the mass of , we use equation 1, we get

Molar mass of = 17 g/mol

the answer to your question is c

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