Question

How many grams of ethane gas (C2H6) are in a 12.7 liter sample at 1.6 atmospheres and 24°C? Show all work used to solve this problem.

Answer 1

Answer on Question #58141 - Chemistry - Other

Task:

How many grams of ethane gas $(C_2H_6)$ are in a 12.7 liter sample at 1.6 atmospheres and $24^{\circ}C$? Show all work used to solve this problem.

Solution:

The ideal gas law is the equation of state of a hypothetical ideal gas. It is a good approximation to the behavior of many gases under many conditions. The ideal gas law is written as:

$$pV = nRT$$

where:

- $p$ is the pressure of the gas;

- $V$ is the volume of the gas;

- $n$ is the amount of substance of gas;

- $R$ is the ideal, or universal, gas constant, equal to the product of the Boltzmann and the Avogadro constant. The gas constant value is $R = k_b \cdot N_a = 8.3144598(48) \, \text{J mol}^{-1} \, \text{K}^{-1}$;

- $T$ is the temperature of the gas.

The chemical amount (n) (in moles) is equal to the mass (m) (in grams) divided by the molar mass (M) (in grams per mole):

$$n = \frac{m}{M}$$

By replacing $n$ with $m / M$ we get:

$$pV = \frac{m}{M}RT$$

We perform algebraic transformation so that on one side of the sign = was an unknown quantity, and on the other - all known.

$$m = \frac{pVM}{RT}$$

Calculate the molar mass (M) of ethane gas $(C_2H_6)$:

$$M(C_2H_6) = 12 \cdot 2 + 1 \cdot 6 = 30 \, (\text{grams/mole}) = 30 \cdot 10^{-3} \, (\text{kg/mole});$$

Convert some magnitude:

$$T = t^{\circ} + 273.15 = 24 + 273.15 = 297.15 \, \text{K};$$

Since 1 liter = $10^{-3}$ m³, the 12,7 liter = $12.7 \times 10^{-3}$ m³;

Since 1 atm = 101 325 Pa, the 1.6 atm = 162120 Pa.

The result is:

M = $30 \times 10^{-3}$ kg/mole;

p = 162120 Pa;

T = 297.15 K;

V = $12.7 \times 10^{-3}$ m³;

R = 8.314 J mol⁻¹ K⁻¹.

Then,

$$m = \frac{p V M}{R T} = \frac{162120 \times 12.7 \times 10^{-3} \times 30 \times 10^{-3}}{8.314 \times 297.15} = 0.025 (kg) = 25 (g).$$

Answer:

$m(C_2H_6) = 25g.$

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