What amount of #Zn# will be required to produce #H_2# by its action on dilute #H_2SO_4# which will completely react with the oxygen, produced by heating #30g# of #KClO_3#?

1 Answer
Jun 25, 2018

Answer:

  • #n("Zn")=0.651 color(white)(l) "mol"#
  • #m("Zn") = n("Zn") * M("Zn") = 42.56 color(white)(l) g#

Explanation:

Start with the only known quantity in this question

#m("KClO"_3) = 30 color(white)(l) g#

#n("KClO"_3) = 30 color(white)(l) g * (1 color(white)(l) "mol")/(138.55 color(white)(l) "g")=0.217 color(white)(l) "mol"#

The balanced equation of the decomposition of potassium perchlorate

#color(navy)(2) color(white)(l) "KClO"_3 (s) stackrel(Delta)(to) 2 color(white)(l) "KCl" (s) + color(purple)(3) color(white)(l) "O"_2 (g)#

suggests the stoichiometric relationship

  • #color(navy)(2) color(white)(l) "mol" color(white)(l) "KClO"_3# decomposes to produce #color(purple)(3) color(white)(l) "mol" color(white)(l) "O"_2 (g)#

Hence the amount of oxygen required for the complete combustion of the unknown amount of #"H"_2 (g)# produced would be

#n("O"_2) = 0.217 color(white)(l) "mol" color(white)(l) "KClO"_3 * (color(purple)(3) color(white)(l) "mol" color(white)(l) "O"_2 )/(color(navy)(2) color(white)(l) "mol" color(white)(l) "KClO"_3)=0.326 color(white)(l) "mol" color(white)(l) "O"_2 #

Oxygen reacts with hydrogen by the equation

# color(purple)(1) color(white)(l) "O"_2 (g) + color(navy)(2) color(white)(l) "H"_2 (g)stackrel("*")(to) 2 color(white)(l) "H"_2"O" (g)#

at a #color(purple)(1): color(navy)(2)# ratio, meaning that the combustion would consume

#n("H"_2) = 0.326 color(white)(l) "mol" color(white)(l) "O"_2 * (color(navy)(2) color(white)(l) "H"_2 )/(color(purple)(1) color(white)(l) "O"_2)=0.651 color(white)(l) "mol" color(white)(l) "H"_2#

of hydrogen. All these #"H"_2# came from the reaction between #"Zn"# and dilute sulfuric acid #"H"_2 "SO"_4# as seen in the following equation

#color(navy)(1) color(white)(l) "Zn"(s) + "H"_2"SO"_4 (aq) to color(navy)(1) color(white)(l) "H"_2 (g) + "ZnSO"_4 (aq)#

where for each mole of #"H"_2 (g)# produced, #color(navy)(1) color(white)(l) "mol"# of #"Zn"# is consumed. That is:

#n("Zn") = n("H"_2) = 0.651 color(white)(l) "mol"#

Hence the mass of #"Zn"#

#m("Zn") = n("Zn") * M("Zn") = 42.56 color(white)(l) g#