Question

A mixture containing KClO3,K2CO3,KHCO3, and KCl was heated, producing CO2,O2, and H2O gases according to the following equations:2KClO3(s)?2KCl(s)+3O2(g)2KHCO3(s)?K2O(s)+H2O(g)+2CO2(g)K2CO3(s)?K2O(s)+CO2(g)The KCl does not react under the conditions of the reaction. 100.0 g of the mixture produces 1.90 g of H2O, 13.64 of CO2, and 4.10 g of O2. (Assume complete decomposition of the mixture.)1. How many grams of KClO3 were in the original mixture?2.How many grams of KHCO3 were in the original mixture?

Answer 1

Answer:

For 1: The mass of $KClO_3$ in the original mixture is 10.46 g

For 2: The mass of $KHCO_3$ in the original mixture is 21.11 g

Explanation:

We are given:

Mass of water = 1.90 g

Mass of carbon dioxide = 13.64 g

Mass of oxygen = 4.10 g

• For 1:

The given chemical reaction for decomposition of $KClO_3$ follows:

$2KClO_3\rightarrow 2KCl+3O_2$

By Stoichiometry of the reaction:

$(3\times 32)=96g$ of oxygen is produced when $(2\times 122.5)=245g$ of potassium chlorate is decomposed.

So, 4.10 g of oxygen will be produced when = $\frac{245}{96}\times 4.10=10.46g$ of potassium chlorate is decomposed.

Amount of $KClO_3$ in the mixture = 10.46 g

• For 2:

The given chemical reaction for decomposition of $KHCO_3$ follows:

$2KHCO_3\rightarrow K_2O+H_2O+2CO_2$

By Stoichiometry of the reaction:

18 g of water is produced when $(2\times 100)=200g$ of potassium bicarbonate is decomposed.

So, 1.90 g of water will be produced when = $\frac{200}{18}\times 1.90=21.11g$ of potassium bicarbonate is decomposed.

Amount of $KHCO_3$ in the mixture = 21.11 g