Question #09a86

1 Answer
Jul 19, 2017

Answer:

(B) #2# #"mol KClO"_3#

Explanation:

We're asked to find the number of moles of #"KClO"_3# that need to decompose to yield #67.2# #"L O"_2# (assuming #100%# yield).

The decomposition reaction of potassium chlorate is

#2"KClO"_3(s) rarr 2"KCl"(s) + 3"O"_2(g)#

At s.t.p., one mole of any (ideal) gas occupied a volume of #22.41# #"L"#, so we can use this to convert from liters of oxygen to moles of oxygen:

#67.2cancel("L O"_2)((1color(white)(l)"mol O"_2)/(22.41cancel("L O"_2))) = color(red)(3.00# #color(red)("mol O"_3#

Now, we'll use the coefficients of the chemical equation to calculate the relative number of moles of #"KClO"_3# that need to decompose:

#color(red)(3.00)cancel(color(red)("mol O"_2))((2color(white)(l)"mol KClO"_3)/(3cancel("mol O"_2))) = color(blue)(2# #color(blue)("mol KClO"_3#

Thus, option (B) is correct.