What mass of potassium chlorate (V) must be heated to give 0.001dm^3 of oxygen at 20 degrees celsius and 100 kPa?
2KClO3 (s) ----> 2KCl (s) + 3O2 (g)
This question is so confusing. I don't know if I'm supposed to convert everything to SI units, but then mass is in grams, so how can I convert grams into m^3 for the potassium chlorate
2KClO3 (s) ----> 2KCl (s) + 3O2 (g)
This question is so confusing. I don't know if I'm supposed to convert everything to SI units, but then mass is in grams, so how can I convert grams into m^3 for the potassium chlorate
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Firstly we need work out how many moles of oxygen are present in 0.001 dm^3 of oxygen
PV=nRT
n = PV/RT
n = 100000 Pa x 1x10^-6 m^3 / 8.314 J K-1 mol-1 x 293 K = 4.1x10^-5 mols
O2 and KClO3 are in a 2:3 molar ratio. That means for every 2 moles of KClO3 reacted we produce 3 moles of O2. Thats the same as saying for every 1 mole of KClO3 we react we produce 1.5 moles of O2.
So if we produced 4.1x10^-5 moles of O2 we must have reacted (4.1x10^-5 moles)*2/3 = 2.7x10^-5 moles of KClO3.
n = m / MM
m = n x MM = 2.7x10^-5 mol x 122.55 g/mol = 0.0033 g
Therefore we would have needed to react 0.0033 g of KClO3 to produce 0.001 dm^3 of O2
======= follow up
Kenny B is wrong. Look at my method Kenny.
PV=nRT
n = PV/RT
n = 100000 Pa x 1x10^-6 m^3 / 8.314 J K-1 mol-1 x 293 K = 4.1x10^-5 mols
O2 and KClO3 are in a 2:3 molar ratio. That means for every 2 moles of KClO3 reacted we produce 3 moles of O2. Thats the same as saying for every 1 mole of KClO3 we react we produce 1.5 moles of O2.
So if we produced 4.1x10^-5 moles of O2 we must have reacted (4.1x10^-5 moles)*2/3 = 2.7x10^-5 moles of KClO3.
n = m / MM
m = n x MM = 2.7x10^-5 mol x 122.55 g/mol = 0.0033 g
Therefore we would have needed to react 0.0033 g of KClO3 to produce 0.001 dm^3 of O2
======= follow up
Kenny B is wrong. Look at my method Kenny.
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Using the Ideal Gas Eq'n first calculate the moles
PV = nRT
Hence
n = PV / RT
P is in Pascals : 100,000 Pa
V is in cubic metres (m^3) : 0.001/1000 = 0.000001m^3 = 10^-6 m^3 )
R is the Gas Constant 8,31
T is the temperature in Kelvin 20degrees C = 293 K )
Hence
n = 100,000Pa * 10^-6m^3 / 8,31 * 293K
n = 4.1 x 10^-5 moles (of O2)
From the reaction eq'n above, 4.1 x 10^-5 is equivalent to '3'
So 4.1 x 10^-5 x 2 / 3 = 2.738 x10^-5 moles ( is equivalent to '2') The molar ratio of potassium chlorate (V)
The Mr of KClO3 is ( 39 + 35.5 +(3 x 16) = 122.5
Using the eq'n
moles = mass(g) /Mr
mass(g) = moles * Mr
mass(g) ;; 2.738 x 10^-5 x 122.5 = 3.354 x 10^-3 g (of KClO3)
NB
This is a very small mass, (only three thousands of a gram), but since the volume of O2 is only 1 cm^3, (0.001 dm^3) it is probably a reasonable figure!!!!
PV = nRT
Hence
n = PV / RT
P is in Pascals : 100,000 Pa
V is in cubic metres (m^3) : 0.001/1000 = 0.000001m^3 = 10^-6 m^3 )
R is the Gas Constant 8,31
T is the temperature in Kelvin 20degrees C = 293 K )
Hence
n = 100,000Pa * 10^-6m^3 / 8,31 * 293K
n = 4.1 x 10^-5 moles (of O2)
From the reaction eq'n above, 4.1 x 10^-5 is equivalent to '3'
So 4.1 x 10^-5 x 2 / 3 = 2.738 x10^-5 moles ( is equivalent to '2') The molar ratio of potassium chlorate (V)
The Mr of KClO3 is ( 39 + 35.5 +(3 x 16) = 122.5
Using the eq'n
moles = mass(g) /Mr
mass(g) = moles * Mr
mass(g) ;; 2.738 x 10^-5 x 122.5 = 3.354 x 10^-3 g (of KClO3)
NB
This is a very small mass, (only three thousands of a gram), but since the volume of O2 is only 1 cm^3, (0.001 dm^3) it is probably a reasonable figure!!!!
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First, you need moles:
PV = nRT can be rearranged to give:
n = PV/RT
P = 100 kPa
V = 0.001 dm^3
T = 20C = 293K (use 293)
If your R is in units of kPa-dm^3/deg-mol, then you don;t need to convert anything
Note, too, that a dm^3 is also a L, so you can use kPa-L/deg-mol.
At any rate solve for n then multiply your answer by the molecular mass of O2 (32 g/mole) to get your answer in g.
PV = nRT can be rearranged to give:
n = PV/RT
P = 100 kPa
V = 0.001 dm^3
T = 20C = 293K (use 293)
If your R is in units of kPa-dm^3/deg-mol, then you don;t need to convert anything
Note, too, that a dm^3 is also a L, so you can use kPa-L/deg-mol.
At any rate solve for n then multiply your answer by the molecular mass of O2 (32 g/mole) to get your answer in g.