Question

At liftoff, a space shuttle uses a solid mixture of ammonium perchlorate and aluminum powder to obtain great thrust from the volume change of solid to gas. In the presence of a catalyst, the mixture forms solid aluminum oxide and aluminum trichloride and gaseous water and nitrogen monoxide. a) Write the balanced equation for the reaction, and identify the reducind and oxdizing agents. b) How many total moles of gas are produced when 78.0 kg of ammonium perchlorate reacts with a stoichiometric amount of Al. c) Assuming that one mole of gas occupies 22.4 L, what is the change in volume from this reaction?

Answer 1

a)

reaction

ammonium perchlorate and aluminum powder

the reaciton

Al(s) + NH4ClO4(s) = Al2O3(s) + AlCl3(s) + NO(g) + H2O(g)

balanced: (Al, then Cl, then H, thne O/N)

3 Al(s) + 3 NH4ClO4(s) = Al2O3(s) + AlCl3(s) + 3 NO(g) + 6 H2O(g)

Reduction = species that GAINS electrons

Oxidation = process in which a specie will LOSS electrons

Reducing agent = The species that favors reduction, i.e. it will oxidize in order to reduce another species

Oxidizing agent = The species that favors oxidation, i.e. it will reduce in order to oxidise another species

3 Al(s) + 3 NH4ClO4(s)

Al goes from 0 to +3 in AlCl3 and Al2O3

cl goes from +7 in NH4ClO4 --> NH4+ and ClO4- --> Cl + 2(-2) = -1; Cl = -1 + 8 = +7

to -1 in AlCl3

NH4ClO4 --> contains reducing specie, therefore, oxidizing agent

Al(s) --> contains oxidizing specie, therefore, reducing agent

b)

find total mol when

m = 78 kg of NH4ClO4 burns

mol of NH4ClO = mass/MW = 78 *1000 / 117.4891 = 663.89 mol of NH4ClO react

from

3 Al(s) + 3 NH4ClO4(s) = Al2O3(s) + AlCl3(s) + 3 NO(g) + 6 H2O(g)

3 mol of NH4ClO4 --> (3+6) = 9 mol of gas

1:3

663.89 mol of NH4ClO --> 3*663.89 = 1991.67 mol of gases

c)

1 mol = 22.4L

1991.67 mol --> 22.4*1991.67 = 44613.408 Liters