The mass of 2.75 L of isobutane (C₄H₁₀) gas at STP is 7.13 g
To determine the mass of 2.75 L of isobutane,
First, we will determine the number of moles of the isobutane present at STP.
From the equation of ideal gas, we have that
[tex]PV=nRT[/tex]
∴ [tex]n = \frac{PV}{RT}[/tex]
Where P is the pressure
V is the volume
n is the number of moles / amount
R is ideal gas constant
and T is the temperature
From the question,
V = 2.75 L
At STP
P = 1 atm
R = 0.0821 atm L/mol K
T = 273 K
Putting these values into the equation, we get
[tex]n = \frac{1 \times 2.75}{0.0821 \times 273}[/tex]
n = 0.1227 moles
∴ The number of moles of isobutane present is 0.1227 moles
Now, for the mass
Using the formula
[tex]Number \ of \ moles =\frac{Mass}{Molar \ mass}[/tex]
Then,
Mass = Number of moles × Molar mass
Molar mass of isobutane = 58.12 g/mol
∴ Mass of the isobutane = 0.1227 moles × 58.12 g/mol
Mass of the isobutane = 7.13 g
Hence, the mass of 2.75 L of isobutane (C₄H₁₀) gas at STP is 7.13 g
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