one of the ways of producing Nitrogen Monoxide is to reach ammonia (NH3) gas with Dioxgen.....

Question

this reaction yields Nitrogen Monoxide and water if 65g of NH3 are reached with 145g of dioxygen what volume of nitrogen Monoxide will be generated at 75F and 696 Torr

Answer 1

Your question isn't complete, but one assumes that the reaction in question is the following:

4NH₃ + 5O₂ ==> 4NO + 6H₂O

moles NH₃ = 65 g x 1 mole/17g = 3.82 moles

moles of O₂ = 145 g x 1 mole/32 g = 4.53 moles

Limiting reactant = O₂ because according to 4:5 mole ratio of NH₃:O₂, there is slightly less O₂ than needed

moles of NO generated = 4.53 moles _O2 x 4 moles NO/5 moles O₂ = 3.62 moles NO

Volume of NO:

PV = nRT

Convert T to K and convert torr to atm, then plug and chug.

V = nRT/P = (3.62 mol)(0.0821 Latm/Kmol)(297 K)/0.916 atm

V = 96.4 Liters = 96 Liters (to 2 significant figures)