Question

A sample of kclo3 on decomposition yielded 448 ml of oxygen gas at stp . Then the weight of kclo3 originally taken was

Answer 1

given, volume of oxygen gas = 448 ml at STP

so, number of mole oxygen gas at STP = 448/22400 = 2/100 = 0.02mol

KClO3 decomposes into potassium chloride and oxygen

i.e., $2KClO_3\rightarrow 2KCl+3O_2$

it is clear that 2 moles of KClO3 yielded 3 moles of Oxygen.

or, 1 mole of oxygen is yielded by 2/3 mol of KClO3

or, 0.02mol of oxygen is yielded by 2/3 × 0.02 = 0.04/3 mol of KClO3 .

so the weight of KClO3 = mole of KClO3 × molar mass of KClO3

= 0.04/3 × (122.55 g/mol) [ molar mass of KClO3 = 122.55]

= 1.634g

hence, weight of KClO3 originally taken was was 1.634g

Answer 2

Answer:

hope this will help you guys