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
?
Express your answer using three significant figures.
To solve this problem, we can use the ideal gas law:
PV = nRT
Where:
P = pressure of gas (827 torr)
V = volume of gas (144 L)
n = moles of gas
R = ideal gas constant
T = temperature in Kelvin (21°C = 294 K)
First, let's convert the pressure from torr to atm:
1 atm = 760 torr
Therefore:
827 torr * (1 atm / 760 torr) = 1.08816 atm
Next, let's convert the temperature from Celsius to Kelvin:
21°C + 273 = 294 K
Now we can rearrange the ideal gas law equation to solve for moles of gas:
n = PV / RT
n = (1.08816 atm * 144 L) / (0.0821 L-atm/mol-K * 294 K)
n = 6.354 mol
Since the stoichiometry between CaH2 and H2 is 1:1, we need 6.354 moles of CaH2 to produce that much H2.
Now let's calculate the molar mass of CaH2:
Ca = 40.08 g/mol
H = 1.008 g/mol
Molar mass of CaH2 = 40.08 g/mol + 2(1.008 g/mol) = 42.096 g/mol
Finally, let's convert moles of CaH2 to grams:
6.354 mol * 42.096 g/mol = 267.5 g
Therefore, 267.5 grams of CaH2 are needed to generate 144 L of H2 gas at 827 torr and 21°C.