Question

1) You are given an unknown containing KClO₃.
Following the procedure in your manual, you obtain the following fictional data
Note!

Mass of empty test tube	54.5679 g
Mass of test tube + unknown	54.8523 g
Mass of test tube + unknown + MnO2	54.9786 g
Mass of erlenmeyer flask with residual water	308.81 g
Mass of erlenmeyer flask filled with water	364.23 g
Atmospheric pressure	754.9 mmHg
Temperature of water	38.5 °C
Partial pressure of water at 38.5	37.94 mmHg
Mass of test tube + Residue after heating	54.9032 g
Density of water	.996 g/mL

Calculate:

A) Molar volume of Oxygen
(a) Volume of Oxygen collected	_______________ mL
(b) Partial pressure of dry oxygen	_______________ mmHg
(c) Volume of oxygen corrected to STP	_______________ mL
(d) Mass of oxygen produced	_______________ g
(e) Moles of oxygen gas produced	_______________ mol
(f) Calculated molar volume at STP	_______________ L/mol
B) Per cent KClO3 in unknown
(g) Mass of unknown used	_______________ g
(h) Moles of KClO3	_______________ mol
(i) Mass of KClO3 in unknown	_______________ g
(j) Per cent KClO3 in unknown	_______________ %
C) Residual "KCl??" from UNKNOWN
(k) Mass of residual KCl? produced from UNKNOWN	_______________ g
(l) Theoretical yield of KCl if UNKNOWN is 100% KClO3	_______________ g
(m) % Yield of "KCl" based on k and l above	_______________ %

2) Look up in the handbook or elsewhere....

the actual vapor pressure of water at 38.5 °C	_______________ mmHg

3) Know....
How accurately you will measure the volumes and weights?
How accurately you will be able to measure molar volume, % KClO₃
What is the function of MnO₂, the leveling tank
When do you use the Triple Beam Balance and the Analytical Balance and why.
Then think how would your results be affected IF...:
You started with a wet test tube?
You made an error in one or more of your weighings?
You made an error of 1 cm in using the leveling tank?

Answer 1

A) Molar volume of oxygen

Pick up the relevant data from the table to complete this part. Also, we shall include data that we can compute from the given data.

a	Mass of Erlenmeyer flask with residual water (g)	308.81
b	Mass of Erlenmeyer flask filled with water (g)	364.23
c	Mass of water in the flask (g) = b – a	55.42
d	Atmospheric Pressure (mmHg)	754.9
e	Temperature of water (˚C)	38.5
f	Partial pressure of water at 38.5˚C (mmHg)	37.94
g	Partial pressure of dry oxygen at 38.5˚C (mmHg) = d – f	716.96
h	Temperature of water (K) = e + 273	311.5

Volume of water in flask = volume of displaced water = c/(Density of water at 38.5˚C) = (55.42 g)/(0.996 g mL^-1) = 55.6426 mL (I have kept a few guard digits extra) = (55.6426 mL)*(1 L/1000 mL) = 0.0556426 L

We know that 760 mmHg = 1 atm.

Therefore, partial pressure of dry oxygen gas = (716.96 mmHg)*(1 atm/760 mmHg) = 0.9434 atm (I have kept a few guard digits extra).

Volume of oxygen collected = volume of water displaced = 0.0556426 L.

STP is defined as 1 atm pressure and 273 K temperature. Let V L be the volume of the gas at STP. Therefore, we can use the gas law to write

(0.9434 atm)*(0.0556426 L)/(311.5 K) = (1 atm)*(V L)/(273 K)

=====> V = (0.9434)*(0.0556426)*(273)/(311.5)(1) = 0.0460053

The volume of oxygen gas collected is 0.0460053 ≈ 0.046 L at STP (ans)

We can calculate the moles of oxygen gas collected by employing the equation of state n = PV/RT where P = pressure of the gas; V = volume of the gas; T = temperature of the gas and R = gas constant = 0.082 L-atm/mol.K.

Therefore, n = (1 atm)*(0.046 L)/(0.082 L-atm/mol.K).(273 K) = 2.0548*10^-3 ≈ 2.055*10^-3 mole (ans).

Molar mass of oxygen gas produced (O₂) = 32 g/mol.

Therefore, mass of oxygen gas produced = (2.055*10^-3 mole)*(32 g/1 mole) = 0.06576 g ≈ 0.066 g (ans).

B) Percent KClO₃ IN UNKNOWN

We shall pick out and calculate data that are essential for solving this part of the problem.

a	Mass of empty test tube (g)	54.5679
b	Mass of empty test tube + unknown (g)	54.8523
c	Mass of unknown used (g) = b – a	0.2844

Write down the decomposition of KClO₃ reaction as below:

2 KClO₃ ---------> 2 KCl + 3 O₂

There is a 3:2 molar ratio between O₂ produced and KClO₃ used. We have obtained in part (A) above that 2.055*10^-3 mole O₂ gas was produced.

Therefore, moles KClO₃ used = (2.055*10^-3 mole O₂)*(2 mole KClO₃/3 mole O₂) = 1.37*10^-3 mole (ans).

Molar mass of KClO₃ = 122.55 g/mol.

Therefore, mass of KClO₃ used = (1.37*10^-3 mole)*(122.55 g/1 mole) = 0.167893 g ≈ 0.1679 g (ans).

Percent of KClO₃ in the unknown sample = (0.1679 g/0.2844 g)*100 = 59.0365 ≈ 59.04 (ans).