Question

Calculate the volume of oxygen involved at room temperature and pressure when 0.01 mole of potassium chlorate is decomposed

Answer 1

$2KClO_3\longrightarrow6KCl+3O_2$

From this equation we can create the following relation

$2\space moles \space KClO_3\equiv 24×3\space litres \space of\space O_2$

We therefore calculate the volume for 0.01 moles

$=\frac{0.01×24×3}{2}=0.36\space litres$