[afchick7689]

The problem reads: A mixture of potassium chlorate and manganese dioxide weighing 2.00 g was heated until the reaction was complete. After cooling the mixture was found to weigh 1.60 g. What was the weight of manganese dioxide in the sample? (Note manganese dioxide serves as a catalyst).

So i know that being a catalyst, the manganese dioxide assists the reaction without any of it being lost, and ive derived the chemical equations for both parts of the reaction, but its been a long summer and i cant remember what the math is that needs to be done to calculate the weight of the potassium chlorate and thus the manganese dioxide?? helppp!!

[Mitch]

Why don't you write the chemical equation down for us.

[afchick7689]

KClO₃ + MnO₂+heat=KClO₃ + MnO₂

Is that correct? But then how do you find the weight of the two components when you are given the weight of the mixture? I'm not looking for the answer, just a reminder of how to go about solving this...

thanks!

[xiankai]

u can write the MnO2 above the equation sign, the "heat" is not required btw.

actually, this is a decomposition reaction. in order for mass to be loss, a gas most likely would have been formed. based on the law of conservation of mass, u'll know what is the mass of the gas then.

[afchick7689]

so KClO₃--->(over the arrow: MnO₂)K+ClO₃+ MnO₂

--but I still dont understand how you would find the weight of the potassium chloride, or how much of it is lost during the reaction?

[Donaldson Tan]

(Note manganese dioxide serves as a catalyst).

The mass of MnO2 before the reaction is the same after the reaction.

let mass of MnO2 be X

2KClO₃ -> 2KCl +3 O₂

from the stoichiometric ratio. the number of moles of KClO₃ decomposed = number of moles of KCl formed.

let the number of moles of KCl formed be N

molar mass of KCl = 39.1+35.5 = 74.6g/mol
molar mas of KClO₃ = 39.1 + 3X16 = 87.1g/mol

before heating, mass of mixture = 2.00g
this means: X + 87.1N = 2.00

after heating, mass of mixture = 1.60g
this means: X + 74.6N = 1.60

solve for X

[afchick7689]

OOO ok, it makes so much more sense now! thanks so much1!

--anyone want to check my answer now please?

A mixture of potassium chlorate and manganese dioxide weighing 2.00 g was heated until the reaction was complete.  After cooling the mixture it was found to weigh 1.60 g.  What was the weight of manganese dioxide in the sample?  (Note: manganese dioxide serves as a catalyst)

*A catalyst is not used up in the reaction.
            MnO2 (over arrow)
2KClO3  ? 2KCl + 3O2

Molar mass of KCl = 39.1 + 35.05 = 74.15 g/mol
Molar mass of KClO3 = 39.1 + 35.05 + (16*3) = 122.15 g/mol

# mol of Kcl formed = n
mass of MnO2 (doesn’t change!)= x

before heating, mass = 2.00 g :
x + (122.15 g/mol * n) = 2.00g
after heating, mass = 1.60 g :  
x + (74.15 g/mol * n) = 1.60g

x + 122.15n = 2   &   x + 74.15n = 1.6 ,
therefore x = 2 - 122.15n  &   x = 1.6 - 74.15n

So,  2 - 122.15n = 1.6 - 74.15n
          -1.6   -1.6
.4 - 122.15n = -74.15n
    +122.15 n    +122.15n
.4 = 48n
48    48
.008333 = n
plug it back in...
x + (122.15*.008333)= 2 & x + (74.15*.008333)= 1.6
 x + 1.01791666 = 2    &   x + 0.61791666 = 1.6
x = 0.98208334          &      x = 0.98208334  
*The weight of the manganese dioxide in the sample was 0.98208334g. (W/ SIG FIGS = 0.982g)

[xiankai]

looks good. there's an easier alternative though, but u have learnt how to solve difficult questions using algebra