Ammonia gas (nitrogen trihydride) reacts with diatomic oxygen gas at high temper

Question

Ammonia gas (nitrogen trihydride) reacts with diatomic oxygen gas at high temperatures to form nitrogen monoxide gas and water vapor. a) Write the balanced chemical equation for the reaction. b) Calculate the theoretical yield nitrogen monoxide that forms upon reaction of 266 g of ammonia with 812 g of oxygen gas. Identify the Limiting Reactant and the reactant in excess. c) If the actual yield of nitrogen monoxide is 350. g, calculate the percentage yield of nitrogen monoxide. d) Calculate the mass of excess reactant left over.

Explanation / Answer

a) 4NH3+ 5O2 -------> 4NO + 6H2O balanced equation

b) The reaction suggests 4 mole of NH3 reacts with 5 moles of O2 to produce 4 moles of NO and 6 moles of water.

moles= mass/molar mass, molar masses ( in g/mole): NH3= 17, O2=32, NO= 30, H2O= 18

hence 4*17 gm=68 gm of NH3 reacts with 5*32= 160 gm of O2 to give 4*30=120 gm of NO and 6*18= 108 gm of H2O.

hence theoretical mass ratio of NH3: O2= 68:160 =1:2.35

actual mass ratio of NH3: O2= 266:812= 1:3.05

hence excess is O2. All the NH3 gets reacted and is the limiting reactant. Excess is O2.

68 gm of NH3 gives 120 gm of NO

266 gm of NH3 gives 120*266/68 gm of NO= 469.4 gm of NO

Theoretical yield of NO= 469.4 gm of NO.

actual yield of NO= 350 gm. Percent yield= 100* actual yield/ theoretical yield= 100*350/469.4= 74.56%

68 gm of NH3 requires 160 gm of O2

266 gm of NH3 consumes 266*160/68= 626 gm of O2.

mass of O2 left over= mass of O2 supplied- O2 consumed= 812- 626=186 gm

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.