Question

I am planning on building a space station, roughly the size of a small moon, but totally different than the Death Star. One idea is to refine the metal for this totally new structure in space, in order to make best use of a secret, high-powered energy source that totally has the power to destroy an entire system! The plan would be to bring raw Iron Ore to the space station, refine it into Steel, and use it to build this awesome new space station. Iron Ore (usually magnetite) is 72.4 mass percent Iron. Assume that the Steel produced is a mixture of Iron and Chromium, with 18 mass% of the mixture being Chromium. The amount of Steel required for the space station is 1.5 X 1013 kg, and the boss, who is a really powerful dude that is nothing like Emperor Palpatine, is applying a lot of pressure to have this completely original space station constructed in 1 year.

a. What must the input flowrate of Chromium be to meet the construction schedule of this awesome new weapon with no obvious weaknesses, unlike *other* space stations?

b. The ‘other stuff’ that is in the Iron Ore is Oxygen (molecular oxygen, not diatomic Oxygen gas). How much molecular oxygen will be released by this process, on a molar basis (i.e., moles per day)?

Answer 1

a. 85.6 x 10^3 kg/s

b. 3.93 x 10^11 mol/day

Step-by-step explanation:

a.

First, we calculate the amount of chromium required:

Amount of Chromium = 18% of Amount of steel

Amount of Chromium = (0.18)(1.5 x 10^13 kg)

Amount of Chromium = 2.7 x 10^12 kg

Now, for he mass flow rate of chromium required:

Mass Flow Rate = Amount/Time

Mass Flow Rate = 2.7 x 10^12 kg/(1 yr)(365 days/yr)(24 hr/day)(3600 s/hr)

Mass Flow Rate = 85.6 x 10^3 kg/s

b.

First, we calculate the amount of iron required:

Amount of Iron = 82% of Amount of steel

Amount of Iron = (0.82)(1.5 x 10^13 kg)

Amount of Iron = 1.23 x 10^13 kg

Since, the ore has 72.4% of iron in it. Thus, to get the required amount of iron from ore, we need greater amount of ore, which will be calculated as:

Amount of ore = Amount of Iron/72.4%

Amount of ore = (1.23 x 10^13 kg)/0.724

Amount of ore = 1.7 x 10^13 kg

So, the amount of oxygen in ore will be:

Amount of oxygen = Amount of Ore - Amount of Iron

Amount of Oxygen = 1.7 x 10^13 kg - 1.23 x 10^13 kg

Amount of Oxygen = 0.46 x 10^13 kg

Now, for the no. of moles of oxygen:

No. of Moles = Amount of Oxygen/ Molecular mass

No. of Moles = 0.46 x 10^13 kg/0.032 kg/mol

No. of moles = 14.375 x 10^13 mol

Now, for molar rate:

Molar Rate = No. of moles/Time

Molar Rate = 14.375 x 10^13 mol/(1 yr)(365 days/yr)

Molar Rate = 3.93 x 10^11 mol/day