Calcium hydride, CaH2
, reacts with water to form hydrogen gas:
CaH2(s)+2H2O(l)→Ca(OH)2(aq)+2H2(g)
This reaction is sometimes used to inflate life rafts, weather balloons, and the like, where a simple, compact means of generating H2
is desired.
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 hydrogen gas using the ideal gas law:
PV = nRT
Where:
P = pressure (827 torr)
V = volume (144 L)
n = moles of gas
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature in Kelvin (21 + 273 = 294 K)
n = PV / RT
n = (827 torr * 144 L) / (0.0821 L.atm/mol.K * 294 K)
n = 8.23 moles
Next, we need to use the stoichiometry of the reaction to determine how many moles of CaH2 are required to produce 2 moles of hydrogen gas:
1 mol CaH2 produces 1 mol H2
Therefore,
x mol CaH2 produces 8.23 mol H2
x = 8.23 mol CaH2
Now we can calculate the mass of CaH2 required:
molar mass of CaH2 = 40.08 g/mol + 2 * 1.01 g/mol = 42.1 g/mol
mass = moles * molar mass
mass = 8.23 mol * 42.1 g/mol = 346.23 g
Therefore, 346.23 grams of CaH2 are needed to generate 144 L of H2 gas at 827 torr and 21 °C.