A student mixed 40ml of 0.2m Na2co3 solution with 50 ml f 0.25M cacl2 solution , what is the except number of moles of naclkkw that I will be formed ?

Na2CO3 + CaCl2 ==> CaCO3 + 2NaCl

millimols Na2CO3 = mL x M = estimated 8
mmols CaCl2 = estimated 12.5
So you will form 8 mmols (0.008 mols) CaCO3.

To determine the number of moles of NaCl that will be formed when 40 mL of 0.2 M Na2CO3 solution reacts with 50 mL of 0.25 M CaCl2 solution, we need to use the balanced chemical equation for the reaction between sodium carbonate (Na2CO3) and calcium chloride (CaCl2).

The balanced chemical equation for the reaction is:

Na2CO3 + CaCl2 → 2 NaCl + CaCO3

From the equation, we can see that 1 mole of Na2CO3 reacts with 1 mole of CaCl2 to produce 2 moles of NaCl and 1 mole of CaCO3.

First, let's calculate the number of moles of Na2CO3 and CaCl2:

Moles of Na2CO3 = volume (in L) × concentration (in mol/L)
= 0.04 L × 0.2 mol/L
= 0.008 mol

Moles of CaCl2 = volume (in L) × concentration (in mol/L)
= 0.05 L × 0.25 mol/L
= 0.0125 mol

Next, let's determine the limiting reactant. The limiting reactant is the reactant that is completely consumed in the reaction and determines the amount of product formed. The reactant that produces the fewer moles of product will be the limiting reactant.

From the balanced chemical equation, we see that 1 mole of Na2CO3 reacts with 1 mole of CaCl2 to produce 2 moles of NaCl. Therefore, the theoretical yield of NaCl is twice the number of moles of either Na2CO3 or CaCl2.

In this case, since the molar ratio of Na2CO3 to NaCl is 1:2, the limiting reactant will be Na2CO3, as it produces a smaller number of moles of NaCl.

Therefore, the number of moles of NaCl formed will be twice the number of moles of Na2CO3:

Number of moles of NaCl formed = 2 × moles of Na2CO3
= 2 × 0.008 mol
= 0.016 mol

Hence, the expected number of moles of NaCl formed is 0.016 mol.