why does the mass of parcipitate not change when concentration of cacl2 are increased (0.05, 0.075,0.10mol/l)when reacted with nacl2 ( .05mol/l)

I don't understand the question. There IS no ppt between CaCl2 and NaCl. You may have exceeded the solubility of NaCl or CaCl2. By the way, sodium chloride is NaCl, not NaCl2.

oh im sorry its naso4

The mass of precipitate does not change when the concentration of CaCl2 is increased because the reaction between CaCl2 and NaCl2 does not depend on the concentration of CaCl2 to produce the precipitate.

To understand why, let's consider the reaction between CaCl2 and NaCO3:

CaCl2(aq) + NaCO3(aq) → CaCO3(s) + 2NaCl(aq)

In this reaction, CaCO3 is the precipitate (solid), while NaCl is in aqueous form. The formation of the precipitate depends on the limited availability of the reactants rather than their concentration. The formation of a solid from the reactants occurs once a certain amount of product is formed, known as the stoichiometric amount.

Stoichiometry is based on the ratio of reactants and products in a chemical reaction. According to the balanced equation, for every 1 mole of CaCO3 that forms, 1 mole of CaCl2 is consumed. Therefore, the concentration of CaCl2 does not affect the amount of CaCO3 that can be produced.

To determine the mass of the precipitate, you need to know the stoichiometry of the reaction, the limiting reactant, and the molar mass of the precipitate. The limiting reactant is the one that gets consumed first and determines the amount of product formed. In this case, you mentioned that the concentration of NaCl2 is 0.05 mol/L, which implies that it is in excess, and therefore, CaCl2 is the limiting reactant.

To calculate the mass of the precipitate, you would follow these steps:

1. Determine the moles of the limiting reactant (CaCl2) by multiplying its concentration by the volume used.
2. Use the stoichiometry of the balanced equation to determine the moles of the precipitate (CaCO3) produced.
3. Convert the moles of the precipitate to grams using the molar mass of CaCO3.
4. Finally, calculate the mass of the precipitate.

Remember, the concentration of CaCl2 does not affect the mass of the precipitate as long as there is a sufficient amount of the limiting reactant (CaCl2) to react with the excess reactant (NaCl2).