Question

Calculate the mass of potassium chlorate required to liberate 6.72dm3 of oxygen at STP, molar mass of potassium chorate is 122.5g/mol.

Answer 1

As 6.72 dm^3 = 6.72 L

so , 2kclo3⇒2kcl +302

2{122.5} gives 3 {22.4 L} at stp

so 6.72 L will give

6.72{245}/3{22.4}

=73.5 grams of kclo3