The gas is chlorine gas
Density at STP = 3.16 g per L
1 mole of a gas at STP occupies, 22.4 L
Therefore, it means mass of the gas in 22.4 L that is 1 mol gas will be;
= 3.16 g × 22.4 L
= 70.8 g/mol
≈ 71 g/mol
The identity of the gas would be Chlorine gas, whose molar mass is 71 g/mol
The gas is Cl₂ - Chlorine gas
Stochiometry in Chemistry studies about chemical reactions, especially on quantitative, such as the calculation of volume, mass, amount, which is related to the number of ions, molecules, elements, etc.
In chemical calculations the reactants can be determined, the amount of substances that can be expressed in units of mass, volume, moles or determine a chemical formula for example the levels of substances or hydrate molecular formulas.
The volume per mole of a gas is called the molar volume
[tex]\large{\boxed{\bold{V=n\times Vm}}[/tex]
V = volume of gas, liters
n = mole
Vm = molar volume (liters / mol)
Molar volume depends on conditions:
Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure).
P = pressure, atm
V = volume of gas, liters
n = mole
R = constant = 0.082 L atm / mol K
T = 0 ° C = 273 ° K
So, at STP, Vm is 22.4 liters / mol.
Conditions at T 25 ° C and P 1 atm are stated by RTP (Room Temperature and Pressure). Vm in this condition = 24.4 liters / mol
From here we can input into the ideal gas equation to find an unknown value, molar mass (M) so that we know the gas
The density of a gas is 3.16 g / l at STP
for density of gas, The ideal gas equation becomes
[tex]\displaystyle PV = \frac {M} {m} RT[/tex]
[tex]\displaystyle M = \frac {m} {v} \frac {RT} {P}[/tex]
[tex]\displaystyle M = \rho \frac {RT} {P}[/tex]
[tex]\displaystyle M = 3.16\frac {0.082.273} {1}[/tex]
M = 70.74 g/mole
So the possible gas is Cl₂(chlorine) because the atomic mass of Cl is 35.5
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Keywords: gas, density, STP, molar mass
