Solid potassium chlorate (KClO 3 ) decomposes into potassium chloride and oxygen

Question

Solid potassium chlorate (KClO3) decomposes into potassium chloride and oxygen gas. How many moles of potassium chlorate would need to decompose to supply the minimum amount of oxygen needed to burn 32.0 g of methane?: * A. 4/3 B. 3/2 C. 8/3 D. 32/3 E. 6 Solid potassium chlorate (KClO3) decomposes into potassium chloride and oxygen gas. How many moles of potassium chlorate would need to decompose to supply the minimum amount of oxygen needed to burn 32.0 g of methane?: * A. 4/3 B. 3/2 C. 8/3 D. 32/3 E. 6

Explanation / Answer

1st find the mol of O2 required to burn 32.0 g CH4 completely

CH4 + 2O2   —> CO2 + 2H2O

mol of CH4 = mass / molar mass

= 32.0 / 16.0

= 2 mol

according to above reaction,

mol of O2 required = 2*mol of CH4

= 2*2 mol

= 4.0 mol

So, 4.0 mol of O2 must be produced from KClO3

2KClO3   —> 2KCl + 3O2

mol of KClO3 required = (2/3)*mol of O2

= (2/3)*4.0 mol

= 8/3 mol

Answer: C

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