How much ammonia would form by reaction of 2 mL of 0.10 M NaOH with 100 mL of 0.

Question

How much ammonia would form by reaction of 2 mL of 0.10 M NaOH with 100 mL of 0.1 M NH4+ ion solution? NH4+ + OH- ----> NH3 + H2O.

Please show the work, and explain each step. Thank you, in advance.

Explanation / Answer

NH4+ + OH- ----> NH3 + H2O. now mole of NaOH = molarity * V /1000 = 0.1 * 2 /1000 = 2 * 10^-4 and mole of NH4+ = molarity * V/1000 = 0.1 * 100/1000 = 0.01 since one mole of NH4+ and one mole of OH- gives one mole of NH3 and since mole of OH- is lesser than mole of NH4+, OH- is the limiting reactant and hence mole of NH3 formed = 2 * 10^-4 and weight of NH3 formed = mole * M.wt = 2 * 10^-4 * 17 = 0.0034g

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---

---

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