Based on the balanced equation shown in problem 1, perform the following derivat

Question

Based on the balanced equation shown in problem 1, perform the following derivations: a. Calculate the mass of KClO_3 (s) that would lead to formation of 0.4482 g of oxygen gas. (The molar masses of O_2 are 32.0 amu and KClO_3 (s) is 122.5 amu.) b. Assuming that the original unknown mixture contained KClO_3 (s) and KCl (s), with combined mass of 2.1307 g, calculate the mass of KCl in the unknown mixture. c. Calculate the percent of KClO_3 (s) in the unknown mixture. d. Calculate the percent of KCl (s) in the unknown mixture.

Explanation / Answer

Q2

2KClO3 --> 2KCl + 3O2

a)

mass of KClO3 for 0.4482 g of O2 formation

mol of O2 = mass/MW = (0.4482 )/32 = 0.0140 mol of O2

so, ratio is

2 mol of of KClO3 = 3 mol of O2

2/3 --> 2/3*0.0140 = 0.0093333 mol of KClO3 required for such amount of mass

mass = mol*M;W = 0.0093333*122.5 = 1.1433 g of KClO3

b)

mass of KCl --> Total mass - mass of KClO3 = 2.1307-1.1433 = 0.9874 g are KCl

c)

% of KClO3 in mix = mass of KClO3 / total mass * 100% = 1.1433 / 2.1307 *100% = 53.658% is KClO3

d9

KCl in unkown --> 100-%KClO3 = 100-53.658 = 46.342 %

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