How many grams of KCLO3 must be decomposed to prepare 3.36 litres of oxygen at STP ?
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No of moles = Given mass / Molar mass
The molar mass Of KClO3 =39+35.5+48=122.5
Also No of moles = given volume / 22.4
so, Given mass / Molar mass = Given volume / 22.4
Given mass / 122.5 = 3.36 / 22.4
Given mass= 3.36*122.5/22.4
Mass required = 18.375 g
The molar mass Of KClO3 =39+35.5+48=122.5
Also No of moles = given volume / 22.4
so, Given mass / Molar mass = Given volume / 22.4
Given mass / 122.5 = 3.36 / 22.4
Given mass= 3.36*122.5/22.4
Mass required = 18.375 g
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