what volume of 0.40M calcium chloride solution would be required??????????

Question

Suppose you wanted to add just enough 0.40M calcium chloride solution to 75.0 mL of 0.60M sodium carbonate solution for all of the sodium carbonate and calcium chloride to react,,,,,,,,,

Answer 1

(molarity 1) x (volume 1) = (molarity 2) x (volume 2)

You are given M1 and V1 of the known Na2CO3.
You are given M2 of the CaCl2.
Let V2 = unknown volume of CaCl2

Answer 2

.0299 molar answer-->2.ninety 9 moles of chloride/a hundred liters of answer one million.29 molar calcium chloride 696 ml answer ((.0299 mol Cl)/(L Cl answer))*(.696 L answer)*(one million mol CaCl2)/(2 mol Cl)*(L CaCl2 answer)/(one million.29 mol CaCl2))=8.06 ml CaCl2

Answer 3

first write your balanced equation

CaCl2 + Na2(CO3) = 2NaCl + Ca(CO3)

.075 L * .60 Mol/L = .045 Mol Na2CO3

you need .045 mol of CaCl2
.045 Mols /.40 Mols/L = .1125 L

Answer 4

Na2CO3 + CaCl2 --> CaC3 + 2 NaCl

1) convert volume Na2CO3 to moles
(0.0750 L)(0.60 M) = 0.045 moles Na2CO3

2) convert moles Na2CO3 to moles CaCl2
in this case, they are equal, but they aren't always equal
0.045 moles CaCl2

3) convert moles CaCl2 to volume
(0.045 moles) x (1 L / 0.40 moles) = .11 L = 110 mL