Claculate no. of mole produced of H 2 by using ideal gas equation Ideal gas equa
ID: 699912 • Letter: C
Question
Claculate no. of mole produced of H2 by using ideal gas equation
Ideal gas equation
PV = nRT where, P = atm pressure= 800.5 - 28.3 772.2 mmHg = 1.0160 atm,
V = volume in Liter = 78.9 ml = 0.0789 L
n = number of mole = ?
R = 0.08205L atm mol-1 K-1 =Proportionality constant = gas constant,
T = Temperature in K = 280C = 273.15+ 28 = 301.15 K
We can write ideal gas equation
n = PV/RT
Substitute the value
n = (1.0160X 0.0789)/(0.08205X 30.015) = 0.003244 mole
0.003244 mole of H2 gas produced
According to avogadro's law
At STP 1 mole gas occupy volume = 22.414 L then 0.003244 mole of H2 = 0.003244 X 22.414 = 0.07272 liter
0.07272 liter = 72.72 ml
72.72 ml of H2 produced
Explanation / Answer
A .750 gram sample of iron was reacted with hydrochloric acid as follows: 2Fe + 6HCl = 2FeCl3 + 3H2. The barometric pressure was found to be 800.5 mm Hg and the vapor pressure of water at 28 degrees Celcius is 28.3 mm Hg. The volume of hydrogen is collected in the lab over water was 78.9 ml. Calculate the volume of hydrogen gas produced at STP conditions. Calculate the moles of hydrogen gas produced from the .750 g sample of iron. Please show the work too! I would greatly appreciate it!