If 50.0 g of KCl reacts with 50.0 g of O2 to produce KCLO3 according to the following equation how many grams of KCLO3 will be formed

2.2 grams

Explanation:

2KCl + 3O2 - -> 2KClO3

KCl = 39 + 35.5

= 74.5 g/mol

O = 16 g/mol so O2 is 32 g/mol

We need 3 mols of O2 for every 2 mols of KCl

50 g KCl = 50/74.5 =.671 mols of KCl

we need. 671 * 3/2 = 1.007 mols of O2 = 32 g

Now we have more O2 than we need, just use the. 671 mols of KCl, limiting reagent

now

We get 1 mol KClO3 for every mol of KCl

then we get. 671 mols of KClO3

74.5 + 3*16 = 122.5 grams/mol of KClO3

.671 * 122.5 = 2.2 grams