Taking into account the reaction stoichiometry and the definition of molarity, 0.124 L of a 0.209 M KI solution is needed to completely react with 2.43 g of Cu(NO₃)₂
Reaction stoichiometry
In first place, the balanced reaction is:
2 Cu(NO₃)₂ + 4 KI → 2 CuI + I₂ + 4 KNO₃
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- Cu(NO₃)₂: 2 moles
- KI: 4 moles
- CuI: 2 moles
- I₂: 2 moles
- KNO₃: 4 moles
The molar mass of the compounds is:
- Cu(NO₃)₂: 187.55 g/mole
- KI: 165.9 g/mole
- CuI: 190.45 g/mole
- I₂: 253.9 g/mole
- KNO₃: 101 g/mole
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
- Cu(NO₃)₂: 2 moles× 187.55 g/mole= 375.1 grams
- KI: 4 moles× 165.9 g/mole= 663.6 grams
- CuI: 2 moles× 190.45 g/mole= 380.9 grams
- I₂: 2 moles× 253.9 g/mole= 507.8 grams
- KNO₃: 4 moles× 101 g/mole= 404 grams
Mass of KI required
The following rule of three can be applied: If by reaction stoichiometry 375.1 grams of Cu(NO₃)₂ react with 4 moles of KI, 2.43 grams of Cu(NO₃)₂ react with how many moles of KI?
moles of KI= (2.43 grams of Cu(NO₃)₂× 4 moles of KI)÷ 375.1 grams of Cu(NO₃)₂
moles of KI= 0.0259 moles
Being the molarity the number of moles of solute that are dissolved in a given volume, the volume of KI can be calculated as:
volume= 0.0259 moles÷ 0.209 M
volume= 0.124 L
Finally, 0.124 L of a 0.209 M KI solution is needed to completely react with 2.43 g of Cu(NO₃)₂.