Acetylene (C2H2), an important fuel in welding, is produced in the laboratory when calcium carbide (CaC2) reacts with water: CaC2 (s) + 2 H2O (l) LaTeX: \longrightarrow⟶ C2H2 (g) + Ca(OH)2 (aq) For a sample of acetylene collected over water, total gas pressure (adjusted to barometric pressure) is 909 mmHg and the volume is 1,024 ml. At the temperature of the gas (22°C), the vapor pressure of water is 20.7 mmHg. How many grams of acetylene are collected? Enter to 3 decimal places.

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thomasdecabooter thomasdecabooter · Answer 1 · 2019-09-24T02:05:37+02:00

Answer:

There are 1.287 grams of acetylene collected

Explanation:

Total gas pressure = 909 mmHg

Vapor pressure of water = 20.7 mmHg

Pressure of acetylene = 909 mmHg - 20.7 mmHg = 888.3 mmHg

1mmHg = 1 torr

22 ° C + 273.15 = 295.15 Kelvin

Ideal gas law ⇒ pV = nRT

⇒ with p = pressure of the gas in atm

⇒ with V = volume of the gas in L

⇒ with n = amount of substance of gas ( in moles)

⇒ with R = gas constant, equal to the product of the Boltzmann constant and the Avogadro constant (62.36 L * Torr *K^−1 *mol^−1)

⇒ with T = absolute temperature of the gas (in Kelvin)

888.3 torr * 1.024 L = n * 62.36 L * Torr *K^−1 *mol^−1 * 295.15 K

n = 0.04942 moles of C2H2

Mass of C2H2 = 0.04942 moles x 26.04 g/mole = 1.287 g

There are 1.287 grams of acetylene collected

pstnonsonjoku pstnonsonjoku · Answer 2 · 2022-02-03T13:59:03+01:00

The mass of acetylene is 1.29 g.

The equation of the reaction is;

CaC2 (s) + 2 H2O (l) ----->C2H2 (g) + Ca(OH)2 (aq)

We have the folowing information;

Pressure = 909 mmHg - 20.7 mmHg = 888.3 mmHg or 1.17 atm

Volume= 1,024 ml or 1.024 L

Temperature = 22°C or 295 K

Using the formula;

PV = nRT

n = PV/RT

n = 1.17 atm × 1.024 L/0.082 atmLK-1mol-1 × 295 K

n = 1.198/24.19

n = 0.0495 moles

Mass of acetylene = 0.0495 moles × 26 g/mol = 1.29 g

The mass of acetylene is 1.29 g.

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