Ok, I will guess but chemistry is not my field.
I suppose that we are de- oxidizing this stuff and the difference in mass is O2.
for each chemical equation:
x is moles of O2 in one reaction and Y is in the other
2x KNO3---> 2x KNO2 + x O2
2y NaNO3 --->2y NaO2+ y O2
The lost mass of O2 is
(x+y)(32) = 3.2 - 2.64 = .56 grams of O2 boiled away
so
x + y = .0175 grams of O2 boiled away
The mass of KNO3 = 2x(39+3*16)
= 174 x gm
The mass of NaO3 is 2y*(23+3*16)
= 142 y gm
so
142 x + 174 y = 3.2
so in the end we have these two Na and K equations
x + y = .0175
142 x + 174 y = 3.2
solve for moles of K and Na and then go back and get mass in grams of KNO3 from the moles. that mass *100 /3.2 is your percent KNO3
3.2g of a mixture of kn03 and nan03 was heated to constant weight which was found to be 2.64g. What is the % kn03 in mixture?
Who helps me clearly step by step for me.
Thank you .
2 answers
x + y = .0175 **MOLES** of O2 boiled away
1 answer