What is the theoretical yield (in grams) of Fe(OH)3 in the reaction below when 5.00 mL of 0.150 M FeCl3 are reacted with 6.00 mL of 0.200 M NaOH?

Question

FeCl3+3 NaOH=Fe(OH)3+3 NaCl

Can someone please explain step by step what needs to be done and why?

Jeremiah Adams · Accepted Answer

#m("Fe" ("OH")_3) = 0.0427 color(white)(l) "g"#

Start by finding the quantity available- in #color(darkgreen)(mol)#- for each of the reactants. The number of moles of a particular substance present in a solution #n# equals to the product of the volume of the solution #V# and its concentration within that solution #c#. That is:

#n=c*V#

Keep in mind that #"1" color(white)(l) "M" = 1 color(white)(l) mol * color(navy)("L")^(-1)#; Dimensional analysis is probably what you are looking for.

The question provides multiple quantities implying that one of the two species is in excess. The theoretical yield of #"Fe" "(OH)"_3# is dependent only on the quantity of the limiting reactant. It is thus necessary to find the quantity available for both reactants and determine the limiting species before calculating the amount of #"Fe" "(OH)"_3# produced.

In a similar vein,

The limiting reactant is found by comparing the ratio of the two reactant amounts available for this reaction to the amount given in the equation:

The second ratio states that every one mole of #"NaOH"# is capable of removing #color(darkgreen)(0.333) color(white)(l) mol# of #"FeCl"_3# from the system. However, the first ratio is greater than the second- the theoretical ratio- indicating that the amount of #"FeCl"_3# supplied is greater than the capacity of all #"NaOH"# available such that some #"FeCl"_3# will remain unreacted in the mixture after all #"NaOH"# was converted to #"Fe" ("OH")_3# (and hence #"FeCl"_3# is "in excess".)

#"NaOH"# is thus the limiting reactant and thus determines the quantity of #"Fe" ("OH")_3# produced. Their molar ratio is identical to that of their coefficients in the balanced chemical equation. That is:

#(n("Fe" ("OH")_3))/(n("NaOH"))=1/3#

Therefore

#n("Fe" ("OH")_3)=1/3 * n("NaOH") = 4.00 xx 10^(-4) color(white)(l) mol#

Molar mass of #"Fe" ("OH")_3# equals to the sum of #"amu"# of atoms in one formula unit of this compound. That is

#"M"("Fe" ("OH")_3) = overbrace(55.85)^"Fe" + 3 xx overbrace(16.00)^"O" + 3 xx overbrace(1.008)^"H"=106.87 color(white)(l) "g" * mol^(-1)#

Thus,

#m("Fe" ("OH")_3) = n("Fe" ("OH")_3) * "M"("Fe" ("OH")_3)= 0.0427 color(white)(l) "g"#

Owen Ambrose · Answer

undefined To determine the theoretical yield of Fe(OH)3 in the reaction between FeCl3 and NaOH, we first need to write the balanced chemical equation:

FeCl3 + 3NaOH -> Fe(OH)3 + 3NaCl

Next, we need to calculate the number of moles of FeCl3 and NaOH used in the reaction:

For FeCl3:
moles of FeCl3 = (volume of FeCl3 solution in L) * (molarity of FeCl3 solution)
               = (5.00 mL * 0.001 L/mL) * 0.150 mol/L
               ≈ 0.00075 mol

For NaOH:
moles of NaOH = (volume of NaOH solution in L) * (molarity of NaOH solution)
               = (6.00 mL * 0.001 L/mL) * 0.200 mol/L
               ≈ 0.0012 mol

Since the reaction ratio between FeCl3 and Fe(OH)3 is 1:1, the limiting reactant is FeCl3.

Therefore, the theoretical yield of Fe(OH)3 is equal to the number of moles of FeCl3 used, which is approximately 0.00075 mol.

Now, we can calculate the mass of Fe(OH)3 using its molar mass:

mass of Fe(OH)3 = moles of Fe(OH)3 * molar mass of Fe(OH)3
                 = 0.00075 mol * (55.85 g/mol + 3 * 16.00 g/mol)
                 ≈ 0.00075 mol * 113.85 g/mol
                 ≈ 0.08539 g

Therefore, the theoretical yield of Fe(OH)3 is approximately 0.08539 grams.

What is the theoretical yield (in grams) of Fe(OH)3 in the reaction below when 5.00 mL of 0.150 M FeCl3 are reacted with 6.00 mL of 0.200 M NaOH?

FeCl3+3 NaOH=Fe(OH)3+3 NaCl Can someone please explain step by step what needs to be done and why?

FeCl3+3 NaOH=Fe(OH)3+3 NaCl

Can someone please explain step by step what needs to be done and why?