Question

Mandalorian iron, also known by its Mando'a name of beskar, was an extremely durable iron ore whose only known source was the Outer Rim world of Mandalore and its moon, Concordia.

The battle at The Sarlacc Pit in The Return of the Jedi offers us a rare opportunity to study the composition of the rare and coveted mandalorian armor worn by Boba Fett. In this battle Boba Fett fell into the gaping maw of The Sarlacc, where he would be slowly digested over the course a thousand years. If the molarity of the HCl in The Sarlacc stomach is 0.15M, upon which, over the course of a thousand years, four tons (4000 Liters) is produced, we can then titrate the stomach acid after a thousand years and treat the iron content of the mandalorian armor as an antacid tablet containing a diprotic base, since:

Fe + 2 HCl_(aq) -> Fe²⁺_(aq) + 2 Cl^-_(aq) + H_2(g)

If after 1000 years, it took 21.85 mL of 0.1337 M NaOH to neutralize a 25.00 mL sample of The Sarlacc’s stomach acid, what was the mass % of Iron in Boba Fett’s mandalorian armor.

The last known Imperial medical record reports Boba Fett’s mass as 78.12 kg. The last known sensor scan from Boba Fett’s ship, Slave I, reports 93.36 kg for Boba Fett with his armor on.

You may assume The Sarlaac eats only once a millenia since Jabba the Hutt’s death.

(Please assume that while Boba Fett escaped the Sarlacc Pit, his armor did not.)

Answer 1

Total HCl produced over the course of 1000 year = 4000 litres and Molarity is 0.15M, so we have total moles of HCl = 600 mole.

On titrating left over acid (after digestion) required 21.85 ml of 0.1337 M NaOH for 25 ml of acid

We know, mole of NaOH required = 2.921 mole = mole of HCl.

So, for 4000 litres of acid it would be 467.4 litre of HCl.

Thus, total HCl used up in digestion of Iron = Total HCl produced - total HCl remaining = 600 - 467.4 mole = 132.6 mole of HCl.

From the balanced equation, 2 mole of HCl would digest 1 mole of iron thus, total iron digested = 132.6/2 = 66.3 mole of iron. Thus, total mass of iron = 66.3 * 56 = 3.713 kg.

From the data, we can say mass of armor = 93.36 - 78.12 kg = 15.24 kg

Thus, % of iron = (3.713 / 15.24)*100 = 24.36%