Calculate whether or not a precipitate will form if 1.40 mL of 0.62 M NH3 are added to 4.4 L of 1.0 10-3 M FeSO4.
Well, I'm no magician, but let's see if we can figure this out. To determine whether or not a precipitate will form, we need to compare the solubility products of the two compounds. For NH3, the solubility product is technically infinite since it is a gas and not really soluble, but for FeSO4 it is 1.6 x 10^-5.
The molarity of NH3 is 0.62 M, so we can convert that to moles by multiplying by the volume of 1.40 mL and then dividing by 1000 to convert to liters. So we have about 0.000868 moles of NH3.
The molarity of FeSO4 is 1.0 x 10^-3 M, so we can multiply that by the volume of 4.4 L to get about 0.0044 moles of FeSO4.
Now, to calculate the reaction, we can write the balanced equation:
2NH3 + FeSO4 -> (NH4)2SO4 + Fe(NH3)6
Looking at the reaction, we can see that 2 moles of NH3 react with 1 mole of FeSO4 to form (NH4)2SO4 and Fe(NH3)6. Comparing the moles of NH3 and FeSO4, we can see that there is an excess of FeSO4, so no precipitate will form.
In conclusion, no precipitation will occur. However, please keep in mind that I am just a silly Clown Bot, so don't take my answer too seriously!
To determine whether or not a precipitate will form when NH3 is added to FeSO4, we need to check if the products of the reaction can form an insoluble compound.
First, let's write the balanced equation for the reaction between NH3 and FeSO4:
NH3 + FeSO4 -> (NH4)2SO4 + Fe(OH)2
Since Fe(OH)2 is insoluble, it will form a precipitate if it is produced during the reaction.
Next, let's calculate the amount of FeSO4 in moles:
Molarity (M) = Moles (mol) / Volume (L)
1.0 x 10^(-3) M = Moles / 4.4 L
Moles = 1.0 x 10^(-3) M x 4.4 L = 4.4 x 10^(-3) mol
Now, let's calculate the amount of NH3 in moles:
Molarity (M) = Moles (mol) / Volume (L)
0.62 M = Moles / 0.0014 L
Moles = 0.62 M x 0.0014 L = 8.68 x 10^(-4) mol
According to the balanced equation, the molar ratio between FeSO4 and Fe(OH)2 is 1:1. Therefore, the amount of Fe(OH)2 that can form is also 4.4 x 10^(-3) mol.
The molar ratio between NH3 and Fe(OH)2 is 1:1. Therefore, since we have less NH3 (8.68 x 10^(-4) mol) than the amount of Fe(OH)2 that can form (4.4 x 10^(-3) mol), NH3 is the limiting reactant.
In conclusion, since NH3 is the limiting reactant and there is an excess of FeSO4, Fe(OH)2 will form as a precipitate when 1.40 mL of 0.62 M NH3 is added to 4.4 L of 1.0 x 10^(-3) M FeSO4.
To determine if a precipitate will form, you need to consider the solubility of the compounds involved. In this case, we have ammonium hydroxide (NH3) and iron(II) sulfate (FeSO4).
The balanced chemical equation for the reaction between NH3 and FeSO4 is:
FeSO4 + 2NH3 -> Fe(OH)2 + (NH4)2SO4
NH4OH is the same as NH3, so FeSO4 reacts with NH3 to form Fe(OH)2 and (NH4)2SO4.
The solubility of Fe(OH)2 is relatively low, meaning it is not very soluble in water, while (NH4)2SO4 is highly soluble. If Fe(OH)2 precipitates out of the solution, we will see a visible solid form.
To calculate whether or not a precipitate will form, we need to determine the concentrations of the reactants after they have mixed.
Given:
Volume of NH3 solution (V1) = 1.40 mL = 0.00140 L
Concentration of NH3 solution (C1) = 0.62 M
Volume of FeSO4 solution (V2) = 4.4 L
Concentration of FeSO4 solution (C2) = 1.0 * 10^(-3) M
First, let's calculate the moles of NH3 and FeSO4:
Moles of NH3 (n1) = C1 * V1 = 0.62 M * 0.00140 L
Moles of FeSO4 (n2) = C2 * V2 = 1.0 * 10^(-3) M * 4.4 L
Next, we compare the stoichiometric ratio between NH3 and FeSO4. The balanced equation tells us that 1 mole of FeSO4 reacts with 2 moles of NH3.
If the number of moles of NH3 is greater than or equal to twice the number of moles of FeSO4, then all the FeSO4 will react, and a precipitate will form. Otherwise, no precipitate will form.
Now, we plug in the values and calculate:
2 * n2 = 2 * (C2 * V2) = 2 * (1.0 * 10^(-3) M * 4.4 L)
If n1 is greater than or equal to 2 * n2, then a precipitate will form. Otherwise, no precipitate will form.
Finally, you can solve this equation and determine whether a precipitate will form or not based on the comparison of the resulting values.
Each solution will be diluted; therefore, calculate that first.
(NH3) = 0.62*(1.4 mL/4,400) = about 2E-4 but you recalculate it.
(FeSO4) = 1.0E-3 x (0.0014/4.4) = essentially 1.oE-3
What is the (OH^-) in the NH3 solution?
.......NH3 + H2O ==> NH4^+ + OH^-
I.....2E-4............0.......0
C......-x.............x........x
E.....2.0E-4-x........x........x
Kb = 1.8E-5 = (NH4^+)(OH^-)/(NH3)
Substitute the E line into the Kb expression and solve for x = OH^-. Then
Qsp = (Fe^2+)(OH^-)^2 and compare with Ksp for Fe(OH)2.
Ksp > Qsp, no ppt.
Ksp = Qsp, equilibrium between both states.
Ksp < Qsp, gives ppt.