Calculate the volume of oxygen involved at room temperature and pressure when 0.01 mole of potassium chlorate is decomposed
"2KClO_3\\longrightarrow6KCl+3O_2"
From this equation we can create the following relation
"2\\space moles \\space KClO_3\\equiv 24\u00d73\\space litres \\space of\\space O_2"
We therefore calculate the volume for 0.01 moles
"=\\frac{0.01\u00d724\u00d73}{2}=0.36\\space litres"
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