Calcium Hydride reacts with water to produce hydrogen gas:
CaH2(s) + H2O (l)---> Ca(OH)2 (aq) + H2
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 10.0L of H2 gas if the pressure is 740mmHg are 23 degrees Celsius?
R=62.36 L.mmHg/mol.K
I don't know anything about this R you quote. Work it the long way.
Here's an example. Find mole H2 from PV = nRT
http://www.jiskha.com/science/chemistry/stoichiometry.html
CaH2 + 2H2O = Ca(OH)2 + 2H2
H2: P:.97 atm T: 296K V: 10L
PV=nRT
n=PV/RT
n= (.97*10)/(0.08206*296)
n= .40mol
.4 mol H2 (1molCaH2/2mol H2)(1mol/42g) = 8.4 g CaH2
To determine the number of grams of CaH2 needed to generate 10.0L of H2 gas at 740 mmHg and 23 degrees Celsius, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure (740 mmHg)
V = volume in liters (10.0 L)
n = number of moles of H2
R = gas constant (62.36 L.mmHg/mol.K)
T = temperature in Kelvin (23 + 273 = 296 K)
First, we need to find the number of moles of H2 using the ideal gas law equation. Rearranging the equation, we have:
n = PV / RT
Substituting the values:
n = (740 mmHg) * (10.0 L) / (62.36 L.mmHg/mol.K * 296 K)
Simplifying, we find:
n = 2.5 * 10^-3 moles of H2
Now, let's consider the balanced equation:
CaH2(s) + 2H2O(l) ---> Ca(OH)2(aq) + 2H2(g)
From the balanced equation, we can see that 1 mole of CaH2 produces 2 moles of H2 gas.
Therefore, the moles of CaH2 needed to produce the desired amount of H2 gas is:
(2.5 * 10^-3 moles H2) / (2 moles CaH2) = 1.25 * 10^-3 moles CaH2
Now, let's calculate the molar mass of CaH2:
Ca: 40.08 g/mol
H: 1.01 g/mol
Molar mass of CaH2 = (1 * 40.08 g/mol) + (2 * 1.01 g/mol) = 42.10 g/mol
Finally, we can calculate the mass of CaH2 needed using the molar mass and the number of moles:
Mass of CaH2 = (1.25 * 10^-3 moles CaH2) * (42.10 g/mol) = 0.0526 grams of CaH2
Therefore, approximately 0.0526 grams of CaH2 are needed to generate 10.0L of H2 gas at 740 mmHg and 23 degrees Celsius.
To solve this problem, we need to use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in mmHg)
V = volume (in liters)
n = number of moles
R = ideal gas constant (62.36 L.mmHg/mol.K)
T = temperature (in Kelvin)
First, let's convert the pressure from mmHg to atm since the ideal gas constant is expressed in atm units:
740 mmHg * (1 atm / 760 mmHg) = 0.974 atm
Next, convert the temperature from degrees Celsius to Kelvin:
23 degrees Celsius + 273 = 296 K
Now, rearrange the ideal gas law equation to solve for the number of moles:
n = PV / RT
Substitute the given values into the equation:
n = (0.974 atm) * (10.0 L) / (62.36 L.mmHg/mol.K * 296 K)
Simplify:
n = 0.033 mol
According to the balanced chemical equation, 1 mole of CaH2 produces 1 mole of H2. Therefore, we need the same number of moles of CaH2 as H2. So, we need 0.033 moles of CaH2 to produce 0.033 moles of H2.
Now, let's calculate the molar mass of CaH2:
CaH2 = (40.08 g/mol) + 2(1.01 g/mol) = 42.10 g/mol
Finally, we can calculate the mass of CaH2 required to generate 10.0 L of H2 gas:
Mass = number of moles * molar mass
Mass = 0.033 mol * 42.10 g/mol = 1.3873 g (rounded to four decimal places)
Therefore, approximately 1.3873 grams of CaH2 are needed to generate 10.0 liters of H2 gas at a pressure of 740 mmHg and a temperature of 23 degrees Celsius.