A mixture containing \(\mathrm{KClO}_{3}, \mathrm{K}_{2} \mathrm{CO}_{3}, \mathrm{KHCO}_{3},\) and \(\mathrm{KCl}\) was heated, producing \(\mathrm{CO}_{2}, \mathrm{O}_{2},\) and \(\mathrm{H}_{2} \mathrm{O}\) gases according to the following equations: $$ \begin{array}{l}{2 \mathrm{KClO}_{3}(s) \longrightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)} \\ {2 \mathrm{KHCO}_{3}(s) \longrightarrow \mathrm{K}_{2} \mathrm{O}(s)+\mathrm{H}_{2} \mathrm{O}(g)+2 \mathrm{CO}_{2}(g)} \\\ {\mathrm{K}_{2} \mathrm{CO}_{3}(s) \longrightarrow \mathrm{K}_{2} \mathrm{O}(s)+\mathrm{CO}_{2}(g)}\end{array} $$ The KCl does not react under the conditions of the reaction. If 100.0 g of the mixture produces 1.80 \(\mathrm{g}\) of \(\mathrm{H}_{2} \mathrm{O}, 13.20 \mathrm{g}\) of \(\mathrm{CO}_{2}\) and 4.00 \(\mathrm{g}\) of \(\mathrm{O}_{2},\) what was the composition of the original mixture? (Assume complete decomposition of the mixture.)

From the given balanced chemical equations, we can identify the stoichiometric coefficients (the numbers in front of compounds that balance the equations) for each reaction. We also need the molar mass (grams/mole) of each component. Here are the values: - KClO3: \( 2 KClO_3 \rightarrow 2 KCl + 3 O_2\); Molar mass: 39.1 (K) + 35.5 (Cl) + 3*16 (O) = 122.6 g/mol - KHCO3: \( 2 KHCO_3 \rightarrow K_2 O + H_2 O + 2 CO_2\); Molar mass: 39.1 (K) + 1.01 (H) + 12.01 (C) + 3*16 (O) = 100.12 g/mol - K2CO3: \( K_2 CO_3 \rightarrow K_2 O + CO_2\); Molar mass: 2*39.1 (K) + 12.01 (C) + 3*16 (O) = 138.21 g/mol

We need to determine the moles of each product (O2, H2O, and CO2) formed during the reaction. This is done by dividing the mass of each product by their respective molar masses: H2O: Mass: 1.80g; Molar mass: 18.02 g/mol; Moles: 1.80 / 18.02 = 0.1 mol O2: Mass: 4.00g; Molar mass: 2*16 = 32 g/mol; Moles: 4.00 / 32 = 0.125 mol CO2: Mass: 13.2g; Molar mass: 12.01 (C) + 2*16 (O) = 44.01 g/mol; Moles: 13.2 / 44.01 = 0.3 mol

Using the stoichiometric coefficients and the moles of the products formed, we can calculate the moles of each reactant, and then find their mass: 1. KClO3: Moles of O2 produced: 0.125 mol According to the balanced equation, 2 KClO3 -> 3 O2, so (0.125 * 2) / 3 = 0.0833 mol of KClO3. Mass = (0.0833 mol) * (122.6 g/mol) = 10.21 g 2. KHCO3: Moles of H2O and CO2 produced: 0.1 mol and 0.3 mol According to the balanced equation, 2 KHCO3 -> H2O + 2 CO2, so 0.1 mol H2O and (0.3/2) mol CO2 corresponds to 0.1 mol KHCO3. Mass = (0.1 mol) * (100.12 g/mol) = 10.01 g 3. K2CO3: Moles of CO2 produced: 0.3 mol CO2 from KHCO3 reaction: 0.1*2=0.2 mol (from Step 3) CO2 from K2CO3: 0.1 mol (0.3-0.2 mol) According to the balanced equation, K2CO3 -> CO2, so 0.1 mol CO2 corresponds to 0.1 mol K2CO3. Mass = (0.1 mol) * (138.21 g/mol) = 13.82 g 4. KCl: It doesn't react with other reactants and doesn't form any product. To find the mass of KCl, we will subtract the mass of the other three components from the initial mass of the mixture. Mass of KCl: Initial mass of mixture - (Mass of KClO3 + Mass of KHCO3 + Mass of K2CO3) Mass of KCl: 100g - (10.21g + 10.01g + 13.82g) = 65.96g The composition of the original mixture was: - KClO3: 10.21 g - KHCO3: 10.01 g - K2CO3: 13.82 g - KCl: 65.96 g