Question

A mixture containing KClO3, K2CO3, KHCO3, and KCl was heated, producing CO2,O2, and H2O gases according to the following equations:

2KClO3(s)2KHCO3(s)K2CO3(s)→→→2KCl(s)+3O2(g)K2O(s)+H2O(g)+2CO2(g)K2O(s)+CO2(g)

The KCl does not react under the conditions of the reaction. 100.0 g of the mixture produces 1.70 g of H2O, 12.66 g of CO2, and 4.00 g of O2. (Assume complete decomposition of the mixture.)

You may want to reference (Page) Section 3.7 while completing this problem.

a. How many grams of KClO3 were in the original mixture?

Express your answer to three significant figures.

b. How many grams of KHCO3 were in the original mixture?

c. How many grams of K2CO3 were in the original mixture?

Answer 1

2 KC 10₂ > 2414 + 302 K. (O3 → K₂O + CO2 2KH037 KRO + H2O + 2009 (a) It's clear from above that all of a is formed by the dimociation of Kaloz- - No of males of O, formed = Mass = 4.00g Malar may 31.998 g/mol = 0.125 mol As 3 male O₂ is given by 2 moles K6103 jo No of moles of Kc10° = 0.125 ml x 2 = 0.08333 makes * Mass of kilog in original sample= No of moles x Malay mass * = 0.08333 mal x122.559. mata . = 10.2g b) It's also clear that all of Heo ý formed by - dissociation of KHCO3 No of modes of H₂O = 1.70g - 0.0943658 ml 18.015 gimol Also for every mode of Heo a 2 male, co2 is also produced by disociation of KHCO3. U or No of moles of O2 = 2X0.0943658 = 0.1887316 malo he Mass + 10 produced = Noof moux Molay manje m = 0.1887316 mol x 44.01g|mal = 8.3064 1 mole Hoo is produced by I mole KHCO3 Do No of males' of KHCőzz 0.0943658x2= 0.1887316mol 5 Mass of kHcOz in Oruginal sample = 0.1887316 mol X 100. Visglade = 18.99

- (1) CO is produced from decomposition of KHCO₃ as well as R2(03 1 % Mass of Co, produced from Kz (O3 = Total mass of Cos - Mass of (oz produced from KHCO3 = 12.669-8.300g = 4.3543 No of moles of (0g = 4.354g = 0.0989 mol 94.0lgimol 8 As I mole (O2 is produced from 1 mole K200 o No of moles of K₂ (0₂ = 0.0989 mol A e Mass of Ke (O3 in original sample=0.0989 mel x 1382 g/mol = 13.7g