The preparation of KClO3 follows these equations:
2KMnO4 + 16HCl->2KCl + 2MnCl2 + 5Cl2 (g) + 8H2O
3Cl2 (g) + 6KOH ->KClO3 + 5KCl + 3H2O
1) How many moles of KClO3 are produced by one mole of KMnO4
I think the only link between the two equations is Cl2 but I don't understand how to solve this problem
Thanks for the help
To find out how many moles of KClO3 are produced by one mole of KMnO4, we need to use the given equations and the concept of stoichiometry.
Let's start by examining the stoichiometric ratio between KMnO4 and KClO3. According to the first equation, 2 moles of KMnO4 react to produce 2 moles of KClO3.
So, the ratio is 2 moles of KMnO4 : 2 moles of KClO3.
However, we are given only one mole of KMnO4, so we need to convert the given amount of KMnO4 to moles of KClO3 using this ratio.
Since we have a 1:1 ratio (2 moles of KMnO4 to 2 moles of KClO3), we can say that 1 mole of KMnO4 will produce 1 mole of KClO3.
Therefore, one mole of KMnO4 will produce one mole of KClO3.
So, the answer to the question is that one mole of KMnO4 produces one mole of KClO3.