I was given a question on an assignment and I can't figure it out.
The question is
A student collects 90ml of oxygen gas over water at 25c and a total pressure of 99.3kPa from the decomposition of KClO3.
2KClO3(s)-> 2KCl(s)+ 3O2
a) How many mols of O2 were collected?
b)How many grams of KClO3 was decomposed?
For part a) would I change ml to L and then divide this number by the molar mass of O2 to get my answer?
for part b) I have no idea how to even attempt this any help is greatly appreciated.
on a side not would I change kPa to atm? if so is 99.3kPa*1atm/101kPa = .989 ?
Thank you in advance for any attempts made to help me figure this out.
To solve this problem, we'll need to apply some basic principles of stoichiometry and gas laws. Let's break it down step-by-step:
a) To find the number of moles of oxygen gas collected, you will indeed need to convert the mL of gas to L. Remember that 1 L is equal to 1000 mL. So, in this case, you would divide 90 mL by 1000 to get 0.09 L.
Since we are dealing with gases, we need to use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
First, let's convert the given pressure from kPa to atm. You are correct in using the conversion factor of 1 atm = 101 kPa. So, 99.3 kPa * (1 atm / 101 kPa) = 0.981 atm. You rounded it to 0.989 atm, but it should be 0.981 atm.
Now, let's rearrange the ideal gas law equation to solve for the number of moles (n) of oxygen gas:
n = PV / RT
In this case, we have:
P = 0.981 atm (the converted pressure)
V = 0.09 L (the volume of oxygen gas)
R = 0.0821 L·atm/(K·mol) (the ideal gas constant)
T = 25°C + 273.15 = 298.15 K (note that the temperature must be in Kelvin)
Substituting these values into the equation:
n = (0.981 atm * 0.09 L) / (0.0821 L·atm/(K·mol) * 298.15 K)
Now, simply plug in these values into a calculator to solve for n. The answer will give you the number of moles of oxygen gas collected.
b) To determine how many grams of KClO3 were decomposed, we will use the stoichiometry of the balanced chemical equation: 2KClO3(s) -> 2KCl(s) + 3O2.
From the balanced equation, we can see that 2 moles of KClO3 decompose to produce 3 moles of O2. We already determined the number of moles of O2 collected in part a. Now, we need to use this information to find the moles of KClO3 decomposed.
Given that the ratio of moles of O2 to moles of KClO3 is 3:2, we can set up a proportion:
3 moles of O2 / 2 moles of KClO3 = n moles of O2 / x moles of KClO3
This can be rearranged to solve for x, the moles of KClO3:
x = (n moles of O2 * 2 moles of KClO3) / 3 moles of O2
Substitute the value for n that you found in part a, and calculate x.
Finally, to find the mass of KClO3 decomposed, use the molar mass of KClO3 (from the periodic table) and multiply it by x (in moles). This will give you the grams of KClO3 decomposed.
I hope this helps! Let me know if you have any further questions.