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
  • 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