Question

1) You are given an unknown containing KClO3. Following the procedure in your manual, you obtain the following fictional data Note! For this problem use the given (allbeit incorrect) vapor pressure of water Mass of empty test tube 63.4722 g Mass of test tube + unknown 63.7112 g Mass of test tube + unknown + MnO2 63.7661 g Mass of erlenmeyer flask with residual water 311.00 g Mass of erlenmeyer flask filled with water 350.29 g Atmospheric pressure 759.8 mmHg Temperature of water 27.2 °C Partial pressure of water at 27.2 20.27 mmHg Mass of test tube + Residue after heating 63.6938 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 calculated from moles O2 _______________ mol (i) Mass of KClO3 in unknown calculated from moles O2 _______________ g (j) Per cent KClO3 in unknown _______________ % C) Residual "KCl??" from UNKNOWN Residual "KCl??" is what is left in the tube after heating except for the MnO2 (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 27.2 °C _______________ mmHg WARNING...Do not use this number in problem 1

Answer 1

Mass of test tube + unknown + MnO2 = 63.7661 g

Mass of Erlenmeyer flask with residual water = 311.00 g

Mass of Erlenmeyer flask filled with water = 350.29 g

Atmospheric pressure = 759.8 mmHg

Temperature of water = 27.2 °C

Partial pressure of water at 27.2 = 20.27 mmHg

Mass of test tube + Residue after heating = 63.6938 g

Density of water = 0.996 g/mL

(A) Molar volume of Oxygen

(a) Mass of Erlenmeyer flask with residual water = 311.00 g

Mass of Erlenmeyer flask filled with water = 350.29 g

Mass of water increased = (Mass of Erlenmeyer flask filled with water)-(Mass of Erlenmeyer flask with residual water)

= 350.29 - 311.00 = 39.29 g

Volume of water increased = Mass/density

= 39.29 g/ 0.996 g/ml

= 39.45 ml

As we know, the volume of water increased due to the oxygen atom liberated by KClO₃ decomposition.

Volume increased on water = Volume of oxygen collected = 39.45 ml

(b) Partial pressure of dry oxygen

In this experiment,

Total pressure (P_T) = Partial pressure of water( P_H2O) + Partial pressure of dry oxygen (P_O2)

Total pressure (P_T) = Atmospheric Pressure

P_O2 = P_T - P_H2O

P_O2 = 759.8 mmHg - 20.27 mmHg = 739.53 mmHg

Partial Pressure of dry Oxygen (P₂) = 739.53 mmHg

(c) Volume of oxygen corrected to STP

As we know,

STP,

Pressure (P₁) = 1 bar = 750.06 mmHg

Temperatur (T₁) = 273.15 K

Volume (V₁) = ?

P₂ = 739.53 mmHg

T₂ = 27.2 °C = (27.2+273.15) K = 300.35 K

V₂ = 39.45 ml

Using, P₁V₁/T₁ = P₂V₂/T₂

=> V₁ =P₂V₂T₁/P₁T₂

=> V₁ = (739.53x39.45x273.15)/(750.06x300.35)

V₁ = 35.37 ml

(d) Mass of oxygen produced

PV =nRT

n = PV/RT = (739.53x39.45x10^-3)/(62.36367x300.35) = 1.56x10^-3 mol = 1.56 mmol

m/M = 1.56 mmol

m = 1.56x10^-3 x 32 g/mol = 49.84 mg = 4.84 x 10^-3g

R = 62.36367 L·mmHg·K-1·mol-1 (e) From the (d) Moles of oxygen gas produced = 1.56x10-3mol (f) Calculated molar volume at STP Molar volume at STP = Volume gas at STP / number of moles of gas = 35.37 ml/1.56 mmol = 22.7 L/mol (B)  Percent KClO3 in unknown (g) Mass of unknown used Mass of empty test tube = 63.4722 g Mass of test tube + unknown = 63.7112 g Mass of unknown = (Mass of test tube + unknown) - (Mass of empty test tube) = 63.7112 g - 63.4722 g = 0.239 g (h) Moles of KClO3 calculated from moles O2 Balanced equation of KClO3 decomposition 2KClO3 ----> 2KCl + 3O2 ......................(1) From the equation, 1 mole of O2 produced form = 2/3 moles of KClO3 Hence,  Moles of KClO3 calculated from moles O2 = (2/3)x1.56 = 1.04x10-3mol (j) Percent KClO3 in unknown Molar mass of KClO3 = 122.55 g/mol Mass of KClO3 = 1.04x10-3 x122.55 = 0.1275 g Mass of unknown = 0.2390 g Percentage of KClO3 = (mass of KClO3/mass of unknown)x100 = (0.1275/0.2390)x100 = 53.35% (C) Residual "KCl??" from UNKNOWN Residual "KCl??" is what is left in the tube after heating except for the MnO2 Mass of empty test tube = 63.4722 g Mass of test tube + unknown + MnO2 = 63.7661 g Mass of test tube + unknown = 63.7112 g Mass of test tube + Residue after heating = 63.6938 g Mass of Residue after heating = 63.6938 g - 63.4722 g = 0.2216 g Mass of MnO2 = 63.7661 g - 63.7112 g = 0.0549 g From the equation (1), we can say that no. of moles KClO3 decompose is equal to no. of KCl produced. Hence, No. of moles of KCl present = 1.04x10-3mol Molar mass of KCl = 74.6 g/mol (k) mass of KCl produced = 1.04x10-3molx74.6 g/mol = 0.0776 g Mass of residual =   0.2216 g - 0.0549 g = 0.1667 g (without MnO2 ) (l) If Unknown is 100% yield KClO3 Mass of unknown = 0.2390 g Moles of KClO3 = 0.2390/122.55 = 1.95x10-3 mol Moles of KCl = 1.95x10-3 mol Mass of KCl = 1.95x10-3 x 74.6 = 145.47x10-3 g (m) % yield of KCl = (0.0776 g/0.1667 g)x100 = ​46.55 % (based on k) % yield of KCl = (0.1455/0.2390)x100 = 60.88% (based on l)