The formation of KClO 4 can be explained with the given three sequential reactions. The percentage yield of each individual reaction is given in brackets. Cl 2 (g) + 2KOH(aq) → KCl(aq) + KClO(aq) + H 2 O(l) [92 .1%] 3KClO(aq) → 2KCl(aq) + KClO 3 (aq) [86 .7%] 4KClO 3 (aq) → KCl(aq) + 3KClO 4 (aq) [75 .3%] What mass of KClO 4 will be obtained following above three reaction if a student has started with 966 kg of chlorine gas? Concept Introduction: The yield of sequential reactions is based on yield of each previous reaction also. The percentage yield is calculated as. % yield = The amount of product formed × 100 the amount of product expected Also. Moles is related to mass of a compound as. Moles = mass molar mass The atomic mass of each element ( g/mol) Chlorine is 35.5 Potassium is 39 Hydrogen is 1 Oxygen is 16
The formation of KClO 4 can be explained with the given three sequential reactions. The percentage yield of each individual reaction is given in brackets. Cl 2 (g) + 2KOH(aq) → KCl(aq) + KClO(aq) + H 2 O(l) [92 .1%] 3KClO(aq) → 2KCl(aq) + KClO 3 (aq) [86 .7%] 4KClO 3 (aq) → KCl(aq) + 3KClO 4 (aq) [75 .3%] What mass of KClO 4 will be obtained following above three reaction if a student has started with 966 kg of chlorine gas? Concept Introduction: The yield of sequential reactions is based on yield of each previous reaction also. The percentage yield is calculated as. % yield = The amount of product formed × 100 the amount of product expected Also. Moles is related to mass of a compound as. Moles = mass molar mass The atomic mass of each element ( g/mol) Chlorine is 35.5 Potassium is 39 Hydrogen is 1 Oxygen is 16
Solution Summary: The author explains the formation of KClO_ 4 with the given three sequential reactions. The percentage yield of each individual reaction is given in brackets.
The formation of KClO4 can be explained with the given three sequential reactions. The percentage yield of each individual reaction is given in brackets.
A monochromatic light with a wavelength of 2.5x10-7m strikes a grating containing 10,000 slits/cm. Determine the angular positions of the second-order bright line.
Curved arrows are used to illustrate the flow of electrons. Us
the reaction conditions provided and follow the curved arrow
to draw the resulting structure(s).
Include all lone pairs and charges as appropriate.
H
:I
H
0
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell