Exercise 3: if you know that bleaching powder is prepared according to the followingequationCa(OH)2(s) + Cl2(g)Ca(OCI)C(s) + H₂O(1)Five commercial samples of bleaching powder (Ca(OCI)CI) were analyzed using titrationtechnique according to the following equations.Ca(OCI)Cl(s)OCI (aq) + 21 (aq) + H₂O)12(aq) + starch (aq)1₂. starch (aq) + 2 S₂03² (aq)(Deep blue)Ca²+ + OCT (19)(aq)Mass of bleaching powder(g)0.7000.7350.6910.6530.676Hints:+ CT (aq)12(aq) + Cl(aq) + 2OH(aq)1₂. starch(aq)Using the following data, calculate the percent of available chlorine (C1₂%, wt./wt.,70.9 g/mol) in each powder sample.Mw.t21 +S40,²(colorless)Volume (mL) of 0.1 M Mass ofS₂O3² (mL)Cl₂31.733.930.630.028.6% Cl₂availablemole S₂04² = Volume of S₂0₂2 x Molarity of S₂0₂2-1mole I₂ = X mole S₂0₂²2mole OCI= mole 1₂mole Ca(OCI)Cl = mole OCI-mole Cl₂ = mole Ca(OCI)ClAverageMass of Cl₂ = 1/ x Volume of S₂02x Molarity of S₂042 x Molar mass of Cl₂mass of Cl₂ (9)Cl₂ % available =Weight of sample (g)X 100

Bleaching powder is an inorganic compound with chemical formula Ca(OcL)cL. It is used for bleaching…

Exercise 3: if you know that bleaching powder is prepared according to the following equation Ca(OH)2(s) + Cl2(g) Ca(OCI)C(s) + H₂O (1) Five commercial samples of bleaching powder (Ca(OCI)CI) were analyzed using titration technique according to the following equations. Ca(OCI)CI (s) OCT (aq) + 21 (aq) + H₂O) 12(aq) + starch(aq) 1₂. starch (aq) + 2S₂O² (aq) (Deep blue) Hints: Ca2+ (a + OCT (aq) + CF (aq) (aq) 0.700 0.735 0.691 0.653 0.676 12(aq) + Cl(aq) + 2OH(aq) Using the following data, calculate the percent of available chlorine (Cl₂%, wt./wt., Mw.t = 70.9 g/mol) in each powder sample. 1₂. starch(aq) Mass of bleaching powder Volume (mL) of 0.1 M S₂O3 (mL) Mass of Cl₂ = 1 × 21 +S40,² (colorless) 31.7 33.9 30.6 30.0 28.6 Mass of Cl₂ % Cl₂ available mole S₂0²= Volume of S₂042 x Molarity of S₂04²- mole I₂ = x mole S₂04²- 2 mole OCI = mole 1₂ mole Ca(OCI) Cl = mole OCI mole Cl₂ = mole Ca(OCI)Cl Average x Volume of S₂02x Molarity of S₂042 x Molar mass of Cl mass of Cl₂(g) Cl₂ % available = Weight of sample (9) X 100

Exercise 3: if you know that bleaching powder is prepared according to the following equation Ca(OH)2(s) + Cl2(g) Ca(OCI)C(s) + H₂O (1) Five commercial samples of bleaching powder (Ca(OCI)CI) were analyzed using titration technique according to the following equations. Ca(OCI)CI (s) OCT (aq) + 21 (aq) + H₂O) 12(aq) + starch(aq) 1₂. starch (aq) + 2S₂O² (aq) (Deep blue) Hints: Ca2+ (a + OCT (aq) + CF (aq) (aq) 0.700 0.735 0.691 0.653 0.676 12(aq) + Cl(aq) + 2OH(aq) Using the following data, calculate the percent of available chlorine (Cl₂%, wt./wt., Mw.t = 70.9 g/mol) in each powder sample. 1₂. starch(aq) Mass of bleaching powder Volume (mL) of 0.1 M S₂O3 (mL) Mass of Cl₂ = 1 × 21 +S40,² (colorless) 31.7 33.9 30.6 30.0 28.6 Mass of Cl₂ % Cl₂ available mole S₂0²= Volume of S₂042 x Molarity of S₂04²- mole I₂ = x mole S₂04²- 2 mole OCI = mole 1₂ mole Ca(OCI) Cl = mole OCI mole Cl₂ = mole Ca(OCI)Cl Average x Volume of S₂02x Molarity of S₂042 x Molar mass of Cl mass of Cl₂(g) Cl₂ % available = Weight of sample (9) X 100

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Concept explainers

Stoichiometry

Stoichiometry is a chemical division that evaluates the quantitative content of the chemical process with the help of the reacting molecules or products taking part in the reaction. The term “stoicheion” in “Stoichiometry”, is an old Greek term for "element" whereas “metron” means "measure".

Question

Exercise 3: if you know that bleaching powder is prepared according to the following
equation
Ca(OH)2(s) + Cl2(g)
Ca(OCI)C(s) + H₂O(1)
Five commercial samples of bleaching powder (Ca(OCI)CI) were analyzed using titration
technique according to the following equations.
Ca(OCI)Cl(s)
OCI (aq) + 21 (aq) + H₂O)
12(aq) + starch (aq)
1₂. starch (aq) + 2 S₂03² (aq)
(Deep blue)
Ca²+ + OCT (19)
(aq)
Mass of bleaching powder
(g)
0.700
0.735
0.691
0.653
0.676
Hints:
+ CT (aq)
12(aq) + Cl(aq) + 2OH(aq)
1₂. starch(aq)
Using the following data, calculate the percent of available chlorine (C1₂%, wt./wt.,
70.9 g/mol) in each powder sample.
Mw.t
21 +S40,²
(colorless)
Volume (mL) of 0.1 M Mass of
S₂O3² (mL)
Cl₂
31.7
33.9
30.6
30.0
28.6
% Cl₂
available
mole S₂04² = Volume of S₂0₂2 x Molarity of S₂0₂2-
1
mole I₂ = X mole S₂0₂²
2
mole OCI= mole 1₂
mole Ca(OCI)Cl = mole OCI-
mole Cl₂ = mole Ca(OCI)Cl
Average
Mass of Cl₂ = 1/ x Volume of S₂02x Molarity of S₂042 x Molar mass of Cl₂
mass of Cl₂ (9)
Cl₂ % available =
Weight of sample (g)
X 100

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