P. Calculate the mass of potassium chlorate required to liberate 6.72 dm3 of oxygen at STP

Question

4) solve problems

P. Calculate the mass of potassium chlorate required to liberate 6.72 dm³ of oxygen at STP. Molar mass of KClO₃ is 122.5 g mol^-1

Answer 1

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

2 moles of KClO₃ = 2× 122.5=245g

3 moles of O₂ =3× 22.4=67.2 dm³ 

hence , 245 g of potacium chlorate will liberate 67.2 dm³ of oxygen

As , here  x gram of KClO3 liberate 6.72 dm³  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