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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

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

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 dm 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

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