# Titration problem?

Aug 25, 2017

$H C l = 0.00707$ moles
${H}_{2} O = 2.65 \times {10}^{22}$ water molecules

#### Explanation:

First, we calculate the amount of HCl remaining from the NaOH titration. Then we calculate the amount of iron oxide from the amount of HCl that reacted. Finally, the amount of water is the difference between the mass of the sample and the mass of the iron oxide present in the sample.

NaOH Titration of residual HCl
$14.65 m L \times 1.20 M = X m L \times 1.5 M$ ; XmL = 11.72 mL HCl remaining. Therefore, 40 – 11.72 = 28.28 mL were reacted with the rust.
$28.28 m L \times 1.5 M = 42.42 m m o l H C l$

The balanced reaction for the iron oxide – HCl reaction is:

Fe_2O_3 + 6HCl → 2FeCl_3 + 3H_2O ; so it takes 6 mmol of HCl to react with 1 mmol of $F {e}_{2} {O}_{3}$

$\frac{42.42 m m o l H C l}{6} = 7.07 m m o l F {e}_{2} {O}_{3}$, or $0.00707$ moles
Molecular weight of $F {e}_{2} {O}_{3}$ is 159.7g/mol, so 0.00707mol is 1.129g.
1.89g sample – 1.129g iron oxide = 0.761g water.
0.791g water is $\frac{0.791 g}{18 \text{g/mol}} = 0.0439$ moles of water.
Converting this to molecules with Avogadro's Number:

$6.022 \times {10}^{23} \times 0.0439 = 2.65 \times {10}^{22}$ water molecules

Aug 25, 2017

The amount of $\text{HCl}$ reacted was 0.042 mol; $n = 6$.

#### Explanation:

This is an example of a back titration, in which you add excess $\text{HCl}$ to react with the rust, and then you titrate the excess $\text{HCl}$ with $\text{NaOH}$.

Step 1. Calculate the moles of $\text{HCl}$ added

$\text{Moles of HCl" = 0.0400 color(red)(cancel(color(black)("dm"^3color(white)(l) "HCl"))) × "1.50 mol HCl"/(1 color(red)(cancel(color(black)("dm"^3color(white)(l) "HCl")))) = "0.0600 mol HCl}$

Step 2. Calculate the moles of $\text{NaOH}$

$\text{Moles of NaOH" = "0.014 65" color(red)(cancel(color(black)("dm"^3color(white)(l) "NaOHl"))) × "1.20 mol HCl"/(1 color(red)(cancel(color(black)("dm"^3color(white)(l) "HCl")))) = "0.0176 mol HCl}$

Step 3. Calculate the excess moles of $\text{HCl}$

The equation for the titration is

$\text{NaOH + HCl" → "NaCl + H"_2"O}$

$\text{Excess moles of HCl" = 0.0176 color(red)(cancel(color(black)("mol NaOH"))) × "1 mol HCl"/(1 color(red)(cancel(color(black)("mol NaOH")))) = "0.0176 mol HCl}$

Step 4. Calculate the moles of $\text{HCl}$ reacted

$\text{ Moles reacted" = "initial moles - excess moles" = "0.0600 mol - 0.0176 mol" = "0.042 mol}$

Step 5. Calculate the moles of rust reacted

The equation for the reaction is

$\text{Fe"_2"O"_3·n"H"_2"O" + "6HCl" → "2FeCl"_3 + (n+3)"H"_2"O}$

$\text{Moles of rust" = 0.042 color(red)(cancel(color(black)("mol HCl"))) × "1 mol rust"/(6 color(red)(cancel(color(black)("mol HCl")))) = "0.007 07 mol rust}$

Step 6. Calculate the molar mass of the rust

$\text{Molar mass" = "mass"/"moles" = "1.89 g"/"0.007 07 mol" = "267 g/mol}$

Step 7. Calculate the value of $n$

M_text(r, Fe₂O₃·nH₂O) = M_text(r, Fe₂O₃) + nM_text(r, H₂O)

$267 = 159.69 + 18.02 n$

n = (267 -159.69)/18.02 = 108/18.02 = 6.0 ≈ 6

Thus, there are six molecules of water of hydration per formula unit of the rust.