Calculate the final molarity of iodide anion in the solution. potassium carbonate

Find moles, using molar mass:

2.71 g ZnCl2 @ 136.32 g ZnCl2 / mole = 0.01988 moles Zn+2

find moles

0.350. L @ 0.0560 moles / Litre = 0.0196 moles (CO3) - 2

by the equation

Zn+2 & CO3) - 2 - - > ZnCO3 (s)

0.0196 moles (CO3) - 2 reacts with an equal 0.0196 moles Zn+2,

0.01988 moles Zn+2 - 0.0196 moles Zn+2 reacted = 0.00028 moles Zn+2 remain

find molarity remaining

0.00028 moles Zn+2 / 0.350 litres = 0.000799 Molar

your answer, to 1 sig fig is

0.0008 Molar Zn+2

aka

0.8 milliMolar Zn+2