What volume of a 0.430 M NH4I solution is required to react with 289 mL of a 0.560 M Pb(NO3)2 solution?
This question can be interpreted two ways. I will assume you want to know how much NH4I solution must be added to react with ALL of the Pb(NO3)2?
Pb^2+ + 2I^- ==> PbI2
millimols Pb^2+ = mL x M = ?
millimols I^- necessary to react with all of the Pb^2+ is twice that.
Then mL NH4I needed = millimols/M = ?
To solve this problem, you need to use the balanced chemical equation and stoichiometry to determine the ratio of moles of NH4I to Pb(NO3)2. Then, you can use the molarity and volume of one of the solutions to calculate the volume of the other solution required for the reaction.
First, let's write the balanced chemical equation for the reaction between NH4I and Pb(NO3)2:
2 NH4I + Pb(NO3)2 → 2 NH4NO3 + PbI2
From the balanced equation, we can see that the ratio between NH4I and Pb(NO3)2 is 2:1. This means that for every 2 moles of NH4I, we need 1 mole of Pb(NO3)2.
Next, we need to determine the number of moles of Pb(NO3)2 in 289 mL of the 0.560 M solution. We can use the formula:
moles = molarity × volume (in liters)
moles of Pb(NO3)2 = 0.560 M × 0.289 L = 0.16144 moles
Using the stoichiometry from the balanced equation, we know that 1 mole of Pb(NO3)2 reacts with 2 moles of NH4I. Therefore, 0.16144 moles of Pb(NO3)2 will react with 0.32288 moles of NH4I (0.16144 × 2).
Now, we can calculate the volume of the NH4I solution needed to react with 0.32288 moles. We'll use the formula:
volume = moles / molarity (in liters)
volume of NH4I = 0.32288 moles / 0.430 M = 0.7512 liters
To convert the volume from liters to milliliters, we multiply by 1000:
volume of NH4I = 0.7512 liters × 1000 = 751.2 mL
So, approximately 751.2 mL of the 0.430 M NH4I solution is required to react with 289 mL of the 0.560 M Pb(NO3)2 solution.