Question

calculate the mass and volume of oxygen required at STP to convert 3.2 kg of graphite into co2​

Answer 1

Answer:

Mass of oxygen required .:

8544g = 8.544kg

Volume occupied by oxygen .:

at STP = 5980 litre = 5.98 m3

at RTP = 6408 liter = 6.408 m3

Explanation:

The number of moles of carbon atoms in 3.2kg = 3200g of graphite

= 3200/12 = 267 mol

one atom of carbon require one O2 molecule to form one CO2

Thus one mole would require one mole molecules and thus 267 moles of carbon would react with 267 moles of Oxygen molecules.

Now one mole of Oxygen (O2) has a mass of 32 gram thus the total mass of oxygen reuired = 32 × 267 = 8544g

= 8.544kg

Also one mole of Oxygen occupy 22.4 liter (dm^3) volume at STP and 24 at RTP, so the volume occupied by the oxygen .:

at STP = 5980 litre = 5.98 m3

at RTP = 6408 liter = 6.408 m3

Answer 2

Answer:

Graphite is an Allotrope of Carbon

=> $C_{(s)}$ + $O^2_(g)$ ------> $CO^2 _(g)$

∴ 1 molecule of Graphite requires 1 mole of $O^2$ to give 1 molecule of $CO^2$.

∴ The Mole of Graphite:

⇒ Molar Mass of Carbon(C) = 12

= Mass = 2400 kg

= Moles = Mass/Molar Mass

             = 2400/12 = 200 Moles

∴ If 1 mole of Carbon requires 1 mole of $O^2$ then 200 moles requires = 1 x 200/1 = 200 Moles

⇒ Mass of $O^2$ = (Molar Mass = 32, Moles = 200)

= Mass = Moles x Molar Mass

            = 200 x 32

            = 6400 g

            = 6.4 kg

∴ By using Ideal Gas law which states that 1 Mole of an Ideal gas occupies 22.4 $l$

∴ 200 Moles of $O^2$ occupies = 22.4 x 200 $l$

                                               = 4480 $l$

∴ Volume of O² = 4480 liters ($l$)