4) solve problems

P. Calculate the mass of potassium chlorate required to liberate 6.72 dm^{3} of oxygen at STP. Molar mass of KClO_{3 }is 122.5 g mol^{-1}

**\(\begin{matrix}
2KClO_3 &\rightarrow 2KCl&+3O_2\\
2moles & & 3 moles
\end{matrix}\)**

2 moles of KClO_{3} = 2× 122.5=245g

3 moles of O_{2} =3× 22.4=67.2 dm^{3}

hence , 245 g of potacium chlorate will liberate 67.2 dm^{3} of oxygen

As , here x gram of KClO3 liberate 6.72 dm^{3 } of oxygen at stp

\(\therefore x={245\times 6.72\over 67.2}=24.5 g\)

required mass of potacium chlorate is 24.5 g