Question

Consider the following series of balanced chemical reactions in the production of nitric acid,

HNO3: 4 NH3 + 5 O2 -----------> 4 NO + 6 H2O 2 NO + O2 -----------> 2 NO2 3 NO2 + H2O ------------> 2 HNO3 + NO

If O2 and H2O are not limiting, how many grams of nitric acid, HNO3 (F.M. = 63.02), can be produced from the reaction of 120.0 g of ammonia, NH3 (F.M. = 17.04)?

For your calculation, use the given formula masses. Express your answer using only a number or numbers without text. Report your result in decimal notation and to the proper number of significant figures.

Answer 1

4 NH₃ + 5 O₂ -----------> 4 NO

4 moles of ammonia produce 4 moles of nitric oxide.

so 1 mole of ammonia produces 1 mole of nitri oxide.

number of moles of ammonia used = 120 g/17.04 g/mol

= 7.0422 moles.

so nitric oxide available = 7.0422 moles.

2 NO + O₂ -----------> 2 NO₂

from the equation , 1 mole of nitric oxide yields 1 mole of nitrogen dioxide.

so nitrogen dioxide available = 7.0422 moles.

3 NO₂ + H₂O ------------> 2 HNO₃ + NO

as per the equation 3 moles of NO₂ yields 2 moles of nitric acid.

7.0422 moles of NO₂ yields nitric acid = 7.0422 x 2/3

= 4.69 moles of nitri acid produced

4.69 moles of nitric acid = 4.69 x 63.02 g

= 295.87 g

Answer expressed as 295.87  

( toatal 5 significant figures)