1.Moles of O2 generated 2.Mass of KClO3 in sample 3.Percent KClO3 in sample 4.Av

ID: 229489 • Letter: 1

Question

1.Moles of O2 generated 2.Mass of KClO3 in sample 3.Percent KClO3 in sample 4.Average % KClO3 DATA Apparatus UNKNOWN NUMBER: Trial 1 Trial 2 S.204 2l Masw of empry tube af 151g Mass of tube plus sample before heating o.4I 1.233 Mass of tube plus sample after heatingo 461 Volume of water displaced Water temperature ('C) Barometric pressure o. lo. Sa mL 235'c 230'c 139.5H tor CALCULATIONS Show your calculations in the space provided on the next page Molar Volume of Oxygen Mass of Moles of O, generated 0319 0.342g O.OllA mol 0.0106 mol 21.7 thir 2ll torm 11tor 17.44 tar oxygen generated bu 1. Vapor pressure of water Pressure of O Temperature of O, (K) Calculated volume of O ample at STP 32HmL 209 mL Calculated molar volume of O, at STP Pre-labPreSure or 20.4 L 20L Average molar volume Percent error in molar volume

Explanation / Answer

I will base my answers on the calculations you provided.

The average molar volume of O2 is calculated as 20.0 L/mol.

The accepted value of the molar volume is 22.4 L/mol at STP.

Percent error = (accepted value) – (calculated value)/(accepted value)*100 = (22.4 L/mol) – (20.0 L/mol)/(22.4 L/mol)*100 = 10.7143% 10.71% (ans).

Percent KClO3 in sample

Trial 1

Trial 2

Moles of O2 generated

0.01169 mole

0.01069 mole

Mass of KClO3 in sample **

0.955 g

0.874 g

Percent of KClO3 in sample ***

(0.955 g)/(1.572 g)*100% = 60.75%

(0.874 g)/(1.512 g)*100% = 57.80%

Average percent KClO3

½*(60.75 + 57.80)% = 59.275% 59.28%

**(Take Trail 1 as an example)

The balanced chemical equation for the reaction is

2 KClO3 (s) ---------> 2 KCl (s) + 3 O2 (g)

As per the stoichiometric equation,

2 moles KClO3 = 3 moles O2.

Mole(s) of KClO3 corresponding to 0.01169 mole O2 = (0.01169 mole O2)*(2 moles KClO3/3 moles O2) = 0.00779 mole.

Molar mass of KClO3 = (1*39.0983 + 1*35.453 + 3*15.9994) g/mol = 122.5495 g/mol.

Mass of KClO3 corresponding to 0.00779 mole KClO3 = (0.00779 mole)*(122.5495 g/mol) = 0.9547 g

Mass of sample taken = (mass of tube plus sample before heating) – (mass of empty tube after initial heating) = (16.841 g) – (15.269 g) = 1.572 g