The mass (in g) of potassium chlorate required to supply the proper amount of oxygen needed to burn 117.3 g of methane is 1196.82 g
Methane burns in oxygen to produce carbon (iv) oxide and water according to the equation of the reaction below:
CH₄ + 2O₂ ----> CO₂ + 2H₂O
1 mole of methane requires 2 moles of oxygen for complete combustion
1 mole of methane has a mass of 16 g
moles of methane in 117.3 g = 117.3/16 = 7.33 moles of methane
7.33 moles of methane will require 2 * 7.33 moles of oxygen
7.33 moles of methane will require 14.66 moles of oxygen
The decomposition of potassium chlorate produces oxygen
The equation of the reaction is given below:
2 moles of potassium chlorate produces 3 moles of oxygen
14.66 moles of oxygen will be produced by 14.66 * 2/3 moles of potassium chlorate
14.66 moles of oxygen will be produced by 9.77 moles of potassium chlorate
1 mole of potassium chlorate has a mass of 122.5
9.77 moles of potassium chlorate has a mass of 1196.82 g
Therefore, the mass (in g) of potassium chlorate required to supply the proper amount of oxygen needed to burn 117.3 g of methane is 1196.82 g
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