Question

What mass (in g) of potassium chlorate is decomposed if the yield of oxygen gas for the first reaction is 88.6%, if it is known that all produced oxygen was used to make 23.6 g of carbon dioxide and the yield of the second reaction is 77.8%. Methane (CH4) is in excess. Enter a number in g to 1 decimal place. 2 KClO3 ⟶ 2KCl + 3O2 88.6% CH4 + 2O2 ⟶ CO2 + 2H2O 77.8%

Answer 1

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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