Problem 3.80 4 of 144 Part B Constants Periodic Table How many grams of NO and o

Question

Problem 3.80 4 of 144 Part B Constants Periodic Table How many grams of NO and of H2O form? Enter your answers numerically separated by a comma One of the steps in the commercial process for converting ammonia to nitric acid is the conversion of NH; to NO NH (9) + 502(9)-4NO(9) 6H,0(g) In a certain experiment, 2.00 g of NHj reacts with 3,67 g of 02 Submit Request Answer Part C How many grams of the excess reactant remain after the limiting reactant is completely consumed? Express your answer in grams to three significant figures. Submit Request Answer Next> Provide Feedback

Explanation / Answer

Part b

From the stoichiometry of the reaction

(4*17 =) 68 g NH3 reacts with = (5x32=) 160 g of O2

2 g NH3 reacts with = 160x2/68 = 4.705 g of O2

But we have only 3.67 g of O2

O2 is the limiting reactant

(5*32=) 160 g O2 produces = (4*30=) 120 g of NO

3.67 g O2 produces = 120*3.67/160 = 2.7525 g of NO

(5*32=) 160 g O2 produces = (6*18=) 108 g of H2O

3.67 g O2 produces = 108*3.67/160 = 2.477 g of H2O

Mass of NO produced = 2.7525 g

Mass of H2O produced = 2.477 g

Part C

Excess reactant is NH3

(5*32=) 160 g O2 reacts with = (4*17=) 68 g of NH3

3.67 g O2 reacts with = 68*3.67/160 = 1.55975 g of NH3

Excess reactant NH3 remain = 2 - 1.55975 = 0.44025 g

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

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