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

Question

a) What volume of 0.40M calcium chloride solution would be required?

Answer 1

CaCl2 + Na2CO3 ---> CaCO3 +2NaCl
one to one in mols of each

75 *.6 = x * .4

Answer 2

Thanks

Answer 3

B) what mass of calcium carbonate would be produced assuming a 100% yield?

Answer 4

(75/1000).6 = .045 mol
CaCO3 = 100 g/mol
so
4.5 grams