How many grams of CaH2
are needed to generate 144 L
of H2
gas if the pressure of H2
is 827 torr
at 21 ∘C
?
To solve this problem, we first need to calculate the moles of H2 gas using the ideal gas law formula:
PV = nRT
Where:
P = pressure in atm
V = volume in L
n = moles of gas
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature in Kelvin
First, we need to convert the pressure from torr to atm:
827 torr = 827/760 atm = 1.088 atm
Next, we convert the temperature from Celsius to Kelvin:
21 °C = 21 + 273 = 294 K
Now we can plug the values into the formula:
(1.088 atm)(144 L) = n(0.0821 L.atm/mol.K)(294 K)
n = (1.088 atm * 144 L) / (0.0821 L.atm/mol.K * 294 K)
n ≈ 6.35 moles
Since 1 mole of CaH2 produces 2 moles of H2 gas, we need 6.35 moles / 2 = 3.175 moles of CaH2.
The molar mass of CaH2 is approximately 42.1 g/mol.
Therefore, the mass of CaH2 needed is:
3.175 moles * 42.1 g/mol ≈ 133.8 grams
So, approximately 133.8 grams of CaH2 are needed to generate 144 L of H2 gas at a pressure of 827 torr and a temperature of 21 °C.