What is the mass of 16800 mL of oxygen gas at STP

Due to having STP conditions and given volume you can use this conversion: 1 mol=22.4L at STP. Since you have mL, convert to Liters first. 16800 mL = 16.8 L. You will need molar mass of Oxygen(O2)

16.8 L O2 x(1 mol/22.4 L) x( 31.998 g O2/1 mol) = 24.0 g O2

Alternatively, maybe you did not know 1 mol =22.4 L conversion, you can use PV=nRT will work, with P = 1 atm, R =0.08206 L(atm)/mol(k), and T=273 K, V = 16.8 L. Solve for n(moles), then multiply by molar mass of O2

priyankatutor priyankatutor · Answer 2 · 2021-11-18T14:26:48+01:00

The STP is known as standard temperature pressure. The mass of the 16,800 mL of oxygen is 32 g.

Standard Temperature Pressure:

It is the standard environment for reactions at [tex]\bold{25^oC}[/tex] and 1 atm pressure. At STP 22.4 L of gas is equal to 1 mol.

Given here,

Volume of oxygen = 16800 mL = 22.4 L

The molar mass of the Oxygen molecule is 32 g

Number of moles in 22.4 L of oxygen at STP

[tex]\bold{ n = \dfrac{V}{22.4} } \\\\\bold{ n = \dfrac{22.4}{22.4} }\\\\\bold{ n = 1 mol}[/tex]

Mass of 1 mol Oxygen

[tex]\bold{W = n \times m}\\\\\bold{W = 1 \times 32}\\\\\bold{W = 32 g}[/tex]

Therefore, the mass of the 16,800 mL of oxygen is 32 g.

To know more about molar mass, refer to the link:

https://brainly.com/question/12127540