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 .
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
x + y = .0175 **MOLES** of O2 boiled away
To find the percentage of KNO3 in the mixture, you need to follow these steps:
Step 1: Calculate the weight of NaNO3 in the mixture.
Subtract the weight of the mixture after heating from the initial weight of the mixture.
Weight of NaNO3 = Initial weight of the mixture - Weight of the mixture after heating
Weight of NaNO3 = 3.2g - 2.64g
Step 2: Calculate the weight of KNO3 in the mixture.
Subtract the weight of NaNO3 from the initial weight of the mixture.
Weight of KNO3 = Initial weight of the mixture - Weight of NaNO3
Weight of KNO3 = 3.2g - (3.2g - 2.64g)
Step 3: Calculate the percentage of KNO3 in the mixture.
Divide the weight of KNO3 by the initial weight of the mixture and multiply by 100.
% KNO3 = (Weight of KNO3 / Initial weight of the mixture) * 100
Now let's plug in the values and calculate:
% KNO3 = (0.56g / 3.2g) * 100
% KNO3 = 17.5%
Therefore, the percentage of KNO3 in the mixture is 17.5%.