Using the stoichiometry and the number of moles involved in the reaction, The mass of KClO3 decomposed is 63.73 g.
Let us take it up from the second reaction;
Using stoichiometry , the equation of the reaction is;
CH4 + 2O2 ⟶ CO2 + 2H2O
Number of moles of CO2 produced = mass/molar mass
Actual yield of CO2 = 23.6 * 100/77.8 = 30.33 g
Number of moles in 30.33 g of CO2 = 30.33 g/44 g/mol = 0.69 moles
From the reaction equation;
2 moles of O2 yields 1 mole of CO2
x moles of O2 yields 0.69 moles of CO2
x = 2 * 0.69 /1
x = 1.38 moles of O2.
Mass of O2 used = 1.38 moles * 32 g/mol = 44.16 g
This becomes the actual yield of the first reaction.
Theoretical yield of oxygen = 44.16 g * 100/88.6
= 49.84 g of O2
Number of moles of O2 = 49.84 g/32 g/mol = 1.56 moles
Given the reaction equation;
KClO3 ⟶ 2KCl + 3O2
1 mole of KClO3 yields 3 moles of O2
x moles of O2 yields 1.56 moles of O2
x = 1 * 1.56/3
= 0.52 moles of KClO3
Mass of KClO3 decomposed = 0.52 moles * 122.55 g/mol = 63.73 g of KClO3
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