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?

Jun 28, 2018

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

Explanation:

Start by finding the quantity available- in $\textcolor{\mathrm{da} r k g r e e n}{m o l}$- 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:

$n = c \cdot V$

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 ${\text{Fe" "(OH)}}_{3}$ 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 ${\text{Fe" "(OH)}}_{3}$ produced.

• $n \left({\text{FeCl"_3) = c("FeCl"_3) * V("FeCl}}_{3}\right)$
color(white)(n("FeCl"_3)) = 0.150 color(white)(l) "M" xx 5.00 color(white)(l) color(purple)("ml")
color(white)(n("FeCl"_3)) = 0.150 color(white)(l) mol * color(red)(cancel(color(navy)("L")^color(black)(-1))) xx 10^(-3) color(white)(l) color(red)(cancel(color(navy)("L"))) * color(purple)("ml")^(-1) xx 5.00 color(white)(l) color(purple)("ml")
$\textcolor{w h i t e}{n \left({\text{FeCl}}_{3}\right)} = 7.50 \times {10}^{- 4} \textcolor{w h i t e}{l} \textcolor{\mathrm{da} r k g r e e n}{m o l}$

Similarly,

• $n \left(\text{NaOH") = c("NaOH") * V("NaOH}\right) = 1.2 \times {10}^{- 3} \textcolor{w h i t e}{l} \textcolor{\mathrm{da} r k g r e e n}{m o l}$

Comparing the ratio of the amount of the two reactants available to this reaction to that stated in the equation identifies the limiting reactant:

• (n("FeCl"_3, color(purple)("available")))/(n("NaOH", color(purple)("available"))) = color(purple)(0.625)
• (n("FeCl"_3, color(darkgreen)("from equation")))/(n("NaOH", color(darkgreen)("from equation"))) = 1/3 ~~ color(darkgreen)(0.333)

The second ratio states that every one mole of $\text{NaOH}$ is capable of removing $\textcolor{\mathrm{da} r k g r e e n}{0.333} \textcolor{w h i t e}{l} m o l$ of ${\text{FeCl}}_{3}$ from the system. However, the first ratio is greater than the second- the theoretical ratio- indicating that the amount of ${\text{FeCl}}_{3}$ supplied is greater than the capacity of all $\text{NaOH}$ available such that some ${\text{FeCl}}_{3}$ will remain unreacted in the mixture after all $\text{NaOH}$ was converted to "Fe" ("OH")_3 (and hence ${\text{FeCl}}_{3}$ is "in excess".)

$\text{NaOH}$ 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:

$\left(n \left(\text{Fe" ("OH")_3))/(n("NaOH}\right)\right) = \frac{1}{3}$

Hence

$n \left(\text{Fe" ("OH")_3)=1/3 * n("NaOH}\right) = 4.00 \times {10}^{- 4} \textcolor{w h i t e}{l} m o l$

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

$\text{M"("Fe" ("OH")_3) = overbrace(55.85)^"Fe" + 3 xx overbrace(16.00)^"O" + 3 xx overbrace(1.008)^"H"=106.87 color(white)(l) "g} \cdot m o {l}^{- 1}$

Therefore

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