What volume of a 0.610 M NH4I solution is required to react with 525 mL of a 0.300 M Pb(NO3)2 solution?
Pb(NO3)2 + 2NH4I ==> PbI2 + 2NH4NO3
mols Pb(NO3)2 = M x L = estimated 0.16
mols NH4I needed = estd 2*0.16 = about 0.32
M NH4I = mols NH4I/L NH4I. You know mols and M, solve for L.
Why did the ammonium and the lead nitrate have a meeting? Because they wanted to find the perfect balance of chemistry and drama! Now, let's figure out how much volume of the NH4I solution is needed to react with the Pb(NO3)2 solution.
First, we need to determine the number of moles of Pb(NO3)2 in the 525 mL solution. We can do this by multiplying the volume (in liters) by the concentration:
moles Pb(NO3)2 = 0.525 L × 0.300 mol/L = 0.1575 mol Pb(NO3)2
Next, we need to determine the stoichiometry between NH4I and Pb(NO3)2. From the balanced chemical equation, we know that one mole of NH4I reacts with one mole of Pb(NO3)2. So, the number of moles of NH4I required is also 0.1575 mol.
Finally, we can calculate the volume of the NH4I solution. Since the concentration of NH4I is given as 0.610 M, we can plug the moles of NH4I into the equation:
volume NH4I (in liters) = moles NH4I / concentration NH4I
= 0.1575 mol / 0.610 mol/L
= 0.258 L
So, you would need approximately 0.258 liters or 258 mL of the 0.610 M NH4I solution to react with 525 mL of the 0.300 M Pb(NO3)2 solution. That's chemistry in action, my friend!
To determine the volume of the NH4I solution required to react with the Pb(NO3)2 solution, we can use the stoichiometry of the balanced chemical equation between NH4I and Pb(NO3)2.
The balanced chemical equation is:
2 NH4I(aq) + Pb(NO3)2(aq) -> PbI2(s) + 2 NH4NO3(aq)
From the balanced equation, we can see that the stoichiometric ratio between NH4I and Pb(NO3)2 is 2:1. This means that 2 moles of NH4I react with 1 mole of Pb(NO3)2.
First, let's determine the number of moles of Pb(NO3)2 present in 525 mL of the 0.300 M solution:
Moles = concentration x volume
Moles of Pb(NO3)2 = 0.300 M x 0.525 L = 0.1575 moles
Since the stoichiometry ratio is 2:1, we need an equal number of moles of NH4I to react with the Pb(NO3)2. Therefore, we require 0.1575 moles of NH4I.
Now, let's calculate the volume of the NH4I solution required to provide 0.1575 moles:
Volume = Moles / Concentration
Volume of NH4I = 0.1575 moles / 0.610 M = 0.258 L = 258 mL
Therefore, 258 mL of the 0.610 M NH4I solution is required to react with 525 mL of the 0.300 M Pb(NO3)2 solution.
To determine the volume of the NH4I solution required to react with the given volume of Pb(NO3)2 solution, we need to find out the stoichiometry of the reaction between NH4I and Pb(NO3)2.
The balanced chemical equation for the reaction between NH4I and Pb(NO3)2 is:
2NH4I + Pb(NO3)2 → 2NH4NO3 + PbI2
From the balanced equation, we can see that 2 moles of NH4I react with 1 mole of Pb(NO3)2.
First, let's calculate the number of moles of Pb(NO3)2 in the given solution:
Molarity (M) is defined as moles of solute per liter of solution. Therefore, we can use the formula:
moles = molarity × volume (in liters)
Given: Molarity of Pb(NO3)2 = 0.300 M
Volume of Pb(NO3)2 solution = 525 mL = 525/1000 = 0.525 L
moles of Pb(NO3)2 = 0.300 M × 0.525 L = 0.1575 moles
According to the stoichiometry of the reaction, 2 moles of NH4I react with 1 mole of Pb(NO3)2.
Therefore, the number of moles of NH4I required will be half the number of moles of Pb(NO3)2:
moles of NH4I = 0.1575 moles ÷ 2 = 0.07875 moles
Now, let's find the volume of the NH4I solution:
Molarity of NH4I = 0.610 M
Volume of NH4I solution = ?
Using the formula:
moles = molarity × volume (in liters)
0.07875 moles = 0.610 M × volume (in liters)
Solving for volume:
volume (in liters) = 0.07875 moles ÷ 0.610 M = 0.1291 L
Finally, let's convert the volume to milliliters:
volume (in milliliters) = 0.1291 L × 1000 = 129.1 mL
Therefore, approximately 129.1 mL of the 0.610 M NH4I solution is required to react with 525 mL of the 0.300 M Pb(NO3)2 solution.