Final answer:
To react with 2.43 g of Cu(NO₃)₂, 0.124 liters (124 mL) of a 0.209 M KI solution is needed, following the stoichiometric ratio of the chemical reaction.
Step-by-step explanation:
To calculate how many liters of 0.209 M KI solution is needed to completely react with 2.43 g of Cu(NO₃)₂, we must first convert the mass of copper(II) nitrate to moles by using its molar mass. The molar mass of Cu(NO₃)₂ is approximately 187.5 g/mol, so:
moles of Cu(NO₃)₂ = 2.43 g / 187.5 g/mol = 0.01296 mol
According to the balanced chemical equation given, the stoichiometry of Cu(NO₃)₂ to KI is 1:2. Thus, for every mole of Cu(NO₃)₂, two moles of KI are required. So we need:
moles of KI needed = 0.01296 mol Cu(NO₃)₂ * 2 mol KI/mol Cu(NO₃)₂ = 0.02592 mol KI
Now we will convert the moles of KI to liters of the 0.209 M KI solution:
liters of KI solution = moles of KI / Molarity of KI = 0.02592 mol / 0.209 M = 0.124 L
0.124 liters (124 mL) of 0.209 M KI solution is needed to react completely with 2.43 g of Cu(NO₃)₂.