Question

What is the mass of 2.75 L of isobutane (C₄H₁₀) gas at STP? Use atm as the standard pressure unit.

Answer 1

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

$PV=nRT$

∴ $n = \frac{PV}{RT}$

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

$n = \frac{1 \times 2.75}{0.0821 \times 273}$

n = 0.1227 moles

∴ The number of moles of isobutane present is 0.1227 moles

Now, for the mass

Using the formula

$Number \ of \ moles =\frac{Mass}{Molar \ mass}$

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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