To determine the number of grams of CaH2 needed to generate a specific volume of H2 gas, we need to use the ideal gas law equation:
PV = nRT
Where: P = pressure (in atmospheres) V = volume (in liters) n = number of moles R = ideal gas constant (0.0821 L·atm/(mol·K)) T = temperature (in Kelvin)
First, let's convert the given values to the appropriate units: Pressure: 0.995 ATM Volume: 48.0 L Temperature: 32°C = 32 + 273.15 = 305.15 K
Next, we need to calculate the number of moles of H2 gas using the ideal gas law equation. Rearranging the equation, we have:
n = PV / RT
n = (0.995 ATM * 48.0 L) / (0.0821 L·atm/(mol·K) * 305.15 K)
n ≈ 1.956 moles of H2 gas
Now, let's examine the balanced chemical equation for the reaction of CaH2:
CaH2 + 2H2O -> Ca(OH)2 + 2H2
From the equation, we can see that 1 mole of CaH2 produces 2 moles of H2 gas.
Therefore, the number of moles of CaH2 required to produce the calculated amount of H2 gas is also 1.956 moles.
To calculate the mass of CaH2, we need to use its molar mass. The molar mass of CaH2 is 42.09 g/mol.
Mass = moles * molar mass Mass = 1.956 moles * 42.09 g/mol
Mass ≈ 82.34 grams
Therefore, approximately 82.34 grams of CaH2 are needed to generate 48.0 L of H2 gas at a pressure of 0.995 ATM and a temperature of 32°C.