Answer:
To determine the volume of a 0.670 M NH4I solution required to react with 933 mL of a 0.340 M Pb(NO3)2 solution, we can use the concept of stoichiometry and the balanced chemical equation between NH4I and Pb(NO3)2.
First, we need to write the balanced chemical equation for the reaction between NH4I and Pb(NO3)2:
2NH4I + Pb(NO3)2 -> PbI2 + 2NH4NO3
From the balanced equation, we can see that two moles of NH4I react with one mole of Pb(NO3)2 to produce one mole of PbI2 and two moles of NH4NO3.
Next, let's calculate the number of moles of Pb(NO3)2 in the given solution:
Moles of Pb(NO3)2 = concentration * volume
= 0.340 M * 933 mL * (1 L / 1000 mL)
= 0.31662 moles
According to the stoichiometry of the balanced equation, the number of moles of NH4I required to react with Pb(NO3)2 is twice the number of moles of Pb(NO3)2:
Moles of NH4I = 2 * Moles of Pb(NO3)2
= 2 * 0.31662 moles
= 0.63324 moles
Finally, we can calculate the volume of the 0.670 M NH4I solution required using the molarity and moles of NH4I:
Volume of NH4I solution = Moles of NH4I / Molarity of NH4I
= 0.63324 moles / 0.670 M
≈ 0.944 L
Converting this volume to milliliters, we get:
Volume of NH4I solution = 0.944 L * (1000 mL / 1 L)
≈ 944 mL
Therefore, approximately 944 mL of the 0.670 M NH4I solution is required to react with 933 mL of the 0.340 M Pb(NO3)2 solution.
Step-by-step explanation: