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
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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