For the reaction below, when 216g of ICl3 reacts with 61.3g of H2O:

Question

2ICl3 + 3H2O -> ICl + HIO3 + 5HCl

a)which is the limiting reagent?
b)how many grams of HIO3 are formed?
c)how many grams of the nonlimiting reagent are left over?

Answer 1

Convert 216 of ICl3 to mols. mols = grams/molar mass.
Convert 61.3 g H2O to mols the same way.

Using the coefficients in the balanced equation, convert mols ICl3 to mols HIO3.
Using the same procedure, convert mols H2O to mols HIO3.
It is likely that the mols HIO3 will be different; it is obvious that one of them must be wrong. The correct value in limiting reagent problems is ALWAYS the smaller one and the reagent producing that value is the liomitiing reagent.

Using the smaller value for HIO3, convert mols to grams. g = mol x molar mass.

Using mols of the limiting reagent, convert to mols of the non-limiting reagent (use the coefficients as above). That will give you the amount of non-limiting reagent used in the reaction. Convert mols of the "other non-limiting reagent" to grams. Subtract initial amount - amount used to find amount left unreacted.
Post your work if you get stuck.