.969g CaCl2 = 0.00873 moles
1.111g KIO3 = 0.00519 moles
Each mole of CaCl2 requires 2 moles of KIO3, so clearly the KIO3 is the limiting reagent.
So, since each mole of KIO3 produces 1/2 mole of Ca(IO3)2, we get
0.00269 moles of Ca(IO3)2 = 1.051g
if 0.969 g of CaCl2 is mixed with 1.111 g of KlO3 how many grams of solid Ca(IO3)2 can theoretically be made?
CaCl2(aq) + 2KlO3(aq) ===> 2Kcl(aq) + Ca(lO3)2 (s)
1 answer
1 answer