When household ammonia is mixed with bleach, NaOCl, toxic chlorine gas and hydra

Question

When household ammonia is mixed with bleach, NaOCl, toxic chlorine gas and hydrazine, N_2H_4 is produced What volume of 0.75 M NH_3, would react with 100 mL of 0.35 M NaOCl? NH_3 + OCl^- rightarrow Cl_2 + N_2H_4 (unbalanced; basic solution) a 23.5 Ml b, . 100 mL c. 26 mL d 94 mL e 47 mL If 0.050 moles of gaseous HCl is bubbled in 1.0 liter of a solution that is 0.15 M in NH_3 and 0.10 M in NH_4Cl, what will be the pH of the resulting solution? (Assume no volume change after addition of the HCl gas) a. 9.08 b. 4.57 c. 7.05 d. 4.92. e. 9.43 An aqueous solution of nickel (II) acetate is electrolyzed for 3 hours with a 1.8 ampere current. What mass of nickel is produced. a. 7.30 g b. 5.91 g c. 11.5 g d. 0.099 g e. 3.28 g What is the oxidation number of arsenic in [H_2AsO_4]^2-? a. +5 b. +4 c. +2 d. +3 e. +1 The molar solubility of Pbl_2 is 1.52 times 10^-5 M. Calculates the value of K_sp for Pbl_2. a. 3.51 times 10^-9 b. 4.62 times 10^-6 c. 1.40 times 10^-8 d. 1.52 times 10^-8 e none of these. Of Pb^2+, Ag^+, and/or Zn^2+, which could be reduced by Cu? a. Pb^2+ b. Ag^+ c. Zn^+ d. Two of them could be reduced by Cu. e. All of them could be reduced by Cu. For a particular process q = 20 kJ and w = 15kJ. Which of the following statements is true? a. Heat flows from the system to the surroundings. b. Work is done on the system by the surroundings. c. Delta E = +5 kJ. d. Work is done by the system on the surroundings. e. Delta E = -5 kJ.

Explanation / Answer

(27)

NH3 + OCl- ---------------> Cl2 + N2H4

First let us balance this equaiton in basic medium,

(i) Seperate oxidation and reduction half reactions

OCl- ------------> Cl2                 Reduction Half Reaction

NH3 --------------> N2H4           Oxidation Half reaction

(ii) Balance atoms other than O and H

2 OCl- ------------> Cl2

2 NH3 --------------> N2H4

(iii) To balalcne Oxygens in basic medium add OH- ions

2 OCl- ------------> Cl2 + 2 OH-

2 NH3 --------------> N2H4

(iv) To balance Hydrogens in basic medium add H2O molecules

2 OCl- + 2 H2O ------------> Cl2 + 4 OH-

2 NH3   + 2 OH- --------------> N2H4 + 2 H2O

(v) To balance charges add electrons

2 OCl- + 2 H2O + 2e ------------> Cl2 + 4 OH-

2 NH3   + 2 OH- --------------> N2H4 + 2 H2O + 2e

(vi) Add both half reactions to get balanced equation

2 NH3   + 2 OCl- ---------------> N2H4 + Cl2 + 2 OH-

Using the neutralisation formula,

M1V1 / n1 = M2V2 / n2

0.75 * V1 / 1 = 0.35 * 100 / 1

V1 = Volume of NH3 needed = 47 mL

(e)

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

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