A 0.972g sample of a CaCl22H2O/K2C2O4H2O solid salt mixture is dissolved in ~ 150ml of deionized water, previously adjusted to a pH that is basic. The precipitate, after having been filtered and air-dried, has a mass of 0.375 g. The limiting reactant in the salt mixture was later determined to be CaCl2*2H2O. What will be the chemical equation for the reaction?

Question

A 0.972g sample of a CaCl22H2O/K2C2O4H2O solid salt mixture is dissolved in ~ 150ml of deionized water, previously adjusted to a pH that is basic. The precipitate, after having been filtered and air-dried, has a mass of 0.375 g. The limiting reactant in the salt mixture was later determined to be CaCl2*2H2O. What will be the chemical equation for the reaction?

Answer 1

To determine the chemical equation for the reaction, we need to understand the stoichiometry of the reaction. From the given information, we know that the limiting reactant in the salt mixture is CaCl22H2O. This means that there is an excess of K2C2O4H2O in the reaction.

Let's start by writing the chemical formulas for the compounds mentioned in the question:
CaCl2*2H2O - Calcium chloride dihydrate
K2C2O4*H2O - Potassium oxalate monohydrate

Next, let's write the balanced chemical equation for the reaction between CaCl2*2H2O and K2C2O4*H2O. Since CaCl2*2H2O is the limiting reactant, the reaction will proceed based on its stoichiometry.

The balanced chemical equation will be:
CaCl2*2H2O + K2C2O4*H2O -> xCaC2O4*H2O + yKCl + zH2O

To find the coefficients (x, y, and z), we need to use the molar mass and the given mass of the precipitate formed.

1. Calculate the moles of CaCl2*2H2O:
Molar mass of CaCl2*2H2O = (40.08 g/mol + 2 * 35.45 g/mol + 2 * 18.02 g/mol) = 147.02 g/mol
Moles of CaCl2*2H2O = mass / molar mass = 0.972 g / 147.02 g/mol = 0.006616 mol

2. Calculate the moles of K2C2O4*H2O:
Molar mass of K2C2O4*H2O = (2 * 39.10 g/mol + 2 * 12.01 g/mol + 4 * 16.00 g/mol + 18.02 g/mol) = 184.22 g/mol
Moles of K2C2O4*H2O = excess mass / molar mass = (0.375 g - mass of CaC2O4*H2O) / 184.22 g/mol

Given that the excess mass is equal to the mass of the precipitate formed, we can rewrite the equation as:
Moles of K2C2O4*H2O = 0.375 g / 184.22 g/mol

3. Equate the moles of CaCl2*2H2O and K2C2O4*H2O:
0.006616 mol = 0.375 g / 184.22 g/mol

From here, you can solve for the mass of CaC2O4*H2O (x) in the balanced chemical equation and find the values of y and z as well. Then, you can write the final balanced chemical equation using these coefficients.

To determine the chemical equation for the reaction, we need to understand the stoichiometry of the reaction. From the given information, we know that the limiting reactant in the salt mixture is CaCl2*2H2O. This means that there is an excess of K2C2O4*H2O in the reaction.

To determine the chemical equation for the reaction, we need to understand the stoichiometry of the reaction. From the given information, we know that the limiting reactant in the salt mixture is CaCl22H2O. This means that there is an excess of K2C2O4H2O in the reaction.