If potassium chlorate is 80 percent pure ,then 48gram of oxygen would produced from how many gram of potassiumchlorate

Answer 1

2KClO3 ==> 2KCl + 3O2
mols O2 needed = grams/molar mass = 48/32 = about 1.5?
Using the coefficients in the balanced equation, convert mols O2 to mols KClO3. That's 1.5 mols O2 x (2 mols KClO3/3 mols O2) = 1.5 x 2/3 = about 1
So you need 1 mol KClO3 which is grams = mols x molar mass = 1 mol x 122.5 = 122.6 grams IF it were pure. It isn't.
122.6/0.8 = ? if it is 80% pure.

Answer 2

2KClO3=2KCl+3O2.So 245g of KClO3 produces 96g of oxygen.So 122.5g of KClO3 produces 48g of oxygen.This is when it’s 100% pure.But it’s 80% pure.So therefore 152.12 g of KClO3 is required

Answer 3

122.5×0.8

Answer 4

Thank U do more solutions

Answer 5

Simple 100./.means 122.5g
80./. Means 122.5*8o./.