A solution of potassium (k2cro4) when added to a solution of lead(ii) acetate (pb(ch3coo)2 produces a yellow precipitate of lead
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1.06g
Please refer to the image attached.
You first find the number of moles (n) for KClO3 as you have been given the mass. So n = mass/Mr gives you the moles that is 0.022 mol
Next you use the ratio 2:3 as per the equation where it's 2KClO3 ---> 3O2 so you just cross multiply and get the moles for oxygen.
Now that you know the number of moles for oxygen, you can use the same moles equation that you used earlier to find the mass. mass=n×Mr and so you get the answer as 1.06g of oxygen.
