Question

formula?
Q.43 Find the mass of KClO3 required to prepare 3.2 gm of O2.
2KCIO3
KCl + 302
Ans
8.16 gm​

Answer 1

Answer:

Explanation:

first find moles of o2

which is 0.1

divide moles by coefficient

0.1/3

that will be your limiting ratio

and limiting ratio should be multiplied by coefficient for single reactant specie

0.1/3*2

0.067

so moles of kclo3=0.067

mass/122.5=0.067

mass=8.2075 gm

alter method

by poac

moles of kclo3 = moles of kcl

3 moles of kclo3 = 2 moles o2

moles of kclo3 = 0.67 moles o2

alter method

equiavalent of kclo3 = equivalent of o2

moles*f=moles*f

w/122.5 * 6 = 3.2/32*4

w=8.2075 gm