Posted by linda on Saturday, November 24, 2012 at 8:41pm.
6 mols KOH will produce 1 mol KClO3; therefore, 2 mols KOH (1/3 of the 6) must produce 1/3 of the 1 mol KClO3. Right?
is there a formula you use to do this promlem?
There is no formula. There is a process. You convert mols of ANYTHING in the reaction to mols of ANY OTHER reactant OR product by using the coefficients in the balanced equation. For this equation, the coefficients tell you that 1 mol Cl2 will react with 6 mol KOH to produce 1 mol KCLO3 and 5 mol KCl and 3 mols H2O. To make that conversion you simply use the conversion factor which is nothing more than the coefficients. Here is how it works.
Suppose we want to know how many mols H2O will be formed from 6 mols Cl2. That's
6 mols Cl2 x (3 mols H2O/3 mol Cl2) = 6 mols H2O.
How many mols KCl would be formed? That's
6 mols Cl2 x (5 mols KCl/3 mol Cl2) = 10 mols KCl.
How many mols KClO3 would be formed?
That's
6 mols Cl2 x (1 mol KClO3/3 mol Cl2) = 2 mols KClO3.
How many mols KClO3 are formed when 16 mols H2O are formed?
That's
16 mols H2O x (1 mol KClO3/3 mols H2O) = 16/3 = 5.33 mols KClO3.
We can make a formula out of it this way.
mols of what is given x (mols in equation of what we want/mols in equation of what we are given). I hope this helps but please follow up if you have any questions.