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)

.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

To find out how many grams of solid Ca(IO3)2 can be formed, we need to determine which reactant limits the reaction. This can be done by calculating the number of moles of each reactant.

First, let's calculate the number of moles of CaCl2:
Molar mass of CaCl2 = 40.08 g/mol + 2 * 35.45 g/mol = 110.98 g/mol
Number of moles of CaCl2 = mass / molar mass = 0.969 g / 110.98 g/mol ≈ 0.00873 mol

Next, let's calculate the number of moles of KlO3:
Molar mass of KlO3 = 39.1 g/mol + 3 * 16.00 g/mol = 122.1 g/mol
Number of moles of KlO3 = mass / molar mass = 1.111 g / 122.1 g/mol ≈ 0.00910 mol

According to the balanced equation, the ratio of CaCl2 to Ca(IO3)2 is 1:1. This means that the number of moles of Ca(IO3)2 formed will be the same as the number of moles of CaCl2.

Therefore, the number of moles of Ca(IO3)2 that can be theoretically formed is approximately 0.00873 mol.

Now, let's calculate the mass of Ca(IO3)2 using the molar mass of Ca(IO3)2:
Molar mass of Ca(IO3)2 = 40.08 g/mol + 2 * (126.9 g/mol + 3 * 16.00 g/mol) = 391.0 g/mol
Mass of Ca(IO3)2 = number of moles * molar mass = 0.00873 mol * 391.0 g/mol ≈ 3.41 g

Therefore, theoretically, approximately 3.41 grams of solid Ca(IO3)2 can be made.

To determine how many grams of solid Ca(IO3)2 can be made, you need to use stoichiometry, which involves balancing the chemical equation and calculating the molar masses of the reactants and products.

1. Start by balancing the chemical equation:
CaCl2(aq) + 2KlO3(aq) → 2KCl(aq) + Ca(IO3)2(s)

2. Calculate the molar mass of each compound:
- CaCl2: 1 calcium (Ca) atom + 2 chlorine (Cl) atoms = (1 * atomic mass of Ca) + (2 * atomic mass of Cl)
- KlO3: 1 potassium (K) atom + 1 chlorine (Cl) atom + 3 oxygen (O) atoms = (1 * atomic mass of K) + (1 * atomic mass of Cl) + (3 * atomic mass of O)
- KCl: 1 potassium (K) atom + 1 chlorine (Cl) atom = (1 * atomic mass of K) + (1 * atomic mass of Cl)
- Ca(IO3)2: 1 calcium (Ca) atom + 2 iodine (I) atoms + 6 oxygen (O) atoms = (1 * atomic mass of Ca) + (2 * atomic mass of I) + (6 * atomic mass of O)

3. Calculate the number of moles for each reactant:
- Moles of CaCl2 = Mass of CaCl2 (g) / Molar mass of CaCl2 (g/mol)
- Moles of KlO3 = Mass of KlO3 (g) / Molar mass of KlO3 (g/mol)

4. Determine the limiting reactant:
The limiting reactant is the reactant that is completely consumed during the reaction. To identify the limiting reactant, compare the moles of each reactant and their stoichiometric ratios. The reactant with fewer moles or the incorrect stoichiometric ratio will be the limiting reactant.

5. Determine the maximum number of moles of Ca(IO3)2 that can be produced:
The maximum number of moles of Ca(IO3)2 is determined by the stoichiometric ratio between the limiting reactant and Ca(IO3)2. In this case, the stoichiometric ratio is 1:1, meaning 1 mole of Ca(IO3)2 is produced for every mole of the limiting reactant.

6. Convert moles of Ca(IO3)2 to grams:
Multiply the number of moles of Ca(IO3)2 by the molar mass of Ca(IO3)2 to obtain the mass of Ca(IO3)2 in grams.

By following these steps, you can calculate the theoretical mass of solid Ca(IO3)2 that can be made.