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 

Hello! data: p (pressure) = 1.25 atmv (volume) = ? (in l)t (temperature) = 298 kn (number of mols) = 0.900 molesr (gas constant) = 0.082 (atm*l/mol*k)apply the data to the clapeyron equation (ideal gas equation), see: i hope this . =)