[Hemidol]

Many important metals occur as sulfide, arsenide, and antimonide minerals, especially in the Sudbury
mineral complex. The first step in processing these ores involves “roasting” the ore in air to produce the metal
or metal oxide, along with nonmetal oxides that can be serious pollutants if not trapped.
Suppose that you roast 2.00 kg of the mineral polymidite, Ni3S4. The balanced equation for the reaction
is:
2 Ni3S4 + 11 O2 -> 6 NiO + 8 SO2

(a) How many kg of NiO can be produced?
(b) How many kg of the pollutant SO2 can be produced?
(c) How many liters of air at 25°C are required for roasting? Assume that air contains 23% O2 by mass,
and that the density of air at 25°C is 1.2 g/L.
Molar masses: Ni3S4 304.3 g/mol NiO 74.69 g/mol SO2 64.06 g/mol


Here is my work:

(a)
2.00kg Ni₃S₄ * (1000_g / 1kg) * (1 mole Ni₃S₄ / 304.3_g/mole) * (6 mole NiO / 2 mole Ni₃S₄) * (74.69_g/mol / 1 mole NiO) * (1kg / 1000_g) = 1.47kg

(b)

2.00kg Ni₃S₄ * (1000_g / 1kg) * (1 mole Ni₃S₄ / 304.3_g/mole) * (8 mole SO₂ / 2 mole Ni₃S₄) * (64.06_g/mol / 1 mole SO₂) * (1kg / 1000g) = 1.68kg


(c)

Not sure how to approach this problem. Any suggestions are welcome. I will try and reason my way through it if I don't get help within a few hours and see if my logic is correct.


How do a-b look, do they seem logical?

Thakns for reading.

[AWK]

In the same way calculate mass of oxygen needed (if you cannot use the periodic table then the Law of Mass Conservation wil help you), nex mas of air and finally its volume.

[Hemidol]

Mass % = 23g O2 / 100g = 0.23 O2

So...

0.23g O2 * (32g O2 / 1 mole O2) = 7.36g O2

D = M/V

V = 7.36g O2 / 1.2g/L = 6.1 L

Is this correct?

[AWK]

Remember. You start from kilograms of mineral.
At the first step calculate a theoretical mass of oxygen needed.

[Hemidol]

Remember. You start from kilograms of mineral.
At the first step calculate a theoretical mass of oxygen needed.

Reworked...

2 Ni3S4      +     11 O2 -> 6 NiO + 8 SO2
2mol                 11 mol
2x304.3g           11 x 32g
608.6 g               352g
2000 g                (352/608.6) x 2000
                          1156.75g O2

Mass of O2 in 1 L of air = 23% of 1.2g = 0.276g

0.276g of O2 are in 1 litre of air
1156.75g O2 are in (1/0.276) x 1156.75 = 4191.11L