what is the pressure of H2 if the hydrogen gas collected occupies 14L at 300K and was produced upon reaction of 4.5 mol of Al?
PV=nRT
now to n...
2Al+3H2SO4>> Al2(SO4)3 + 3H2
which would indicate 4.5*3/2 moles H2
But what if the acid were HCl?
2Al+ 6HCl>>2AlCl3 + 3H2
moles of H2: 4*3/2 again.
To determine the pressure of H2 gas in this scenario, we can make use of the ideal gas law equation:
PV = nRT
Where:
P = Pressure (in atm)
V = Volume (in liters)
n = Number of moles
R = Ideal gas constant (0.0821 L·atm/(mol·K))
T = Temperature (in Kelvin)
Given:
V = 14 L
T = 300 K
n = 4.5 mol
R = 0.0821 L·atm/(mol·K)
Now, we can plug these values into the ideal gas law equation and solve it for pressure (P):
P * 14 L = 4.5 mol * 0.0821 L·atm/(mol·K) * 300 K
First, let's simplify the right side of the equation:
P * 14 L = 4.5 mol * 0.0821 L·atm/K * 300 K
P * 14 L = 110.92 L·atm
Now, solve for P by dividing both sides of the equation by 14 L:
P = 110.92 L·atm / 14 L
P ≈ 7.92 atm
Therefore, the pressure of H2 gas in this scenario is approximately 7.92 atm.