Consider the thermal decomposition of potassium chlorate.
2 KClO3 (s)-----> 2 KCl (s) + 3 O2 (g)
A mixture containing KCl and KClO3 weighing 1.80 g was heated producing
1.40 x 102 mL of O2 gas at STP. What percent of the original mixture was KClO3?
Is this how I should approach this problem? By using the number of moles?
nmix=nkclo3+nkcl
That might work but it's the long way around AND you don't know mols KCl in the mixture, although you can calculate mols KCl but only after you've worked the problem.
I would use the equation you have to convert mL O2 gas to mols O2 and convert that to mols KClO3. Convert that to grams KClO3. Then
%KClO3 = (g KClO3/mass sample) = ?
Got it. Thank you so much
Yes, you're on the right track! To solve this problem, you'll need to determine the number of moles of KClO3 and KCl in the mixture.
Let's start by calculating the number of moles of O2 gas produced. Since the volume of the O2 gas is given at STP (Standard Temperature and Pressure), you can use the ideal gas law to determine the number of moles.
According to the ideal gas law, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/(mol·K)), and T is the temperature.
At STP, the pressure is 1 atm and the temperature is 273 K. Therefore, we can calculate the number of moles of O2 gas as follows:
nO2 = PV / RT
= (1 atm) * (102 mL) / (0.0821 L·atm/(mol·K) * 273 K)
= 4.00 x 10^-3 mol
According to the balanced equation, the stoichiometric ratio between KClO3 and O2 is 2:3. Therefore, if 2 moles of KClO3 decompose, it would produce 3 moles of O2 gas. From this, we can determine the number of moles of KClO3:
nKClO3 = (3 mol O2 / 2 mol KClO3) * (4.00 x 10^-3 mol O2)
= 6.00 x 10^-3 mol KClO3
Now, let's calculate the number of moles of KCl by subtracting the number of moles of KClO3 from the total number of moles in the mixture:
nKCl = nmix - nKClO3
= 1.80 g / molar mass(KClO3 + KCl) - 6.00 x 10^-3 mol
= (1.80 g / (molar mass(KClO3 + KCl)) - 6.00 x 10^-3 mol
Finally, to find the percentage of KClO3 in the mixture, divide the moles of KClO3 by the total moles of the mixture, then multiply by 100:
Percent KClO3 = (nKClO3 / nmix) * 100
By following these steps, you'll be able to determine the percent of the original mixture that was KClO3.