For the reaction below, when 216g of ICl3 reacts with 61.3g of H2O:
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?
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.
To determine the limiting reagent and the quantities of products formed, we need to use stoichiometry and compare the number of moles of each reactant to the balanced equation.
a) To identify the limiting reagent, we need to find the number of moles of each reactant. We will use the molar masses of ICl3 and H2O to convert their masses into moles.
1. Find the number of moles of ICl3:
Molar mass of ICl3 = 162.23 g/mol
Number of moles of ICl3 = 216 g / 162.23 g/mol
2. Find the number of moles of H2O:
Molar mass of H2O = 18.02 g/mol
Number of moles of H2O = 61.3 g / 18.02 g/mol
Now, compare the mole ratios of ICl3 and H2O in the balanced equation (2:3). The reactant that produces fewer moles of the desired product (HIO3) is the limiting reagent.
b) To calculate the number of grams of HIO3 formed, we need to use the mole ratio between HIO3 and ICl3 from the balanced equation (1:2).
1. Find the number of moles of ICl3 reacting (assuming it is the limiting reagent):
Number of moles of ICl3 = Moles of ICl3 calculated in part a
2. Calculate the number of moles of HIO3 formed:
Number of moles of HIO3 = (Number of moles of ICl3) * (1 mole of HIO3 / 2 moles of ICl3)
3. Find the mass of HIO3 formed:
Molar mass of HIO3 = 175.91 g/mol
Mass of HIO3 = (Number of moles of HIO3) * (Molar mass of HIO3)
c) To determine the grams of the non-limiting reactant left over, we will first calculate the number of moles of the limiting reagent consumed based on the stoichiometry of the balanced equation. Then, we will use the mole ratio of the non-limiting reactant to the limiting reactant to calculate the moles and mass of the non-limiting reactant that remain.
1. Moles of HCl formed:
From the balanced equation, the stoichiometric ratio between ICl3 and HCl is 2:5. Therefore, the number of moles of HCl formed is:
(Number of moles of ICl3 consumed) * (5 moles of HCl / 2 moles of ICl3)
2. Moles of the non-limiting reactant (H2O) reacted:
(Number of moles of H2O) - (Number of moles of H2O consumed based on HCl formation)
3. Find the mass of the non-limiting reactant left over:
Mass of the non-limiting reactant = (Moles of the non-limiting reactant) * (Molar mass of the non-limiting reactant)
By following these steps, you should be able to determine the limiting reagent (a), the grams of HIO3 formed (b), and the grams of the non-limiting reagent remaining (c).