Final answer:
To find the grams of CaH₂ needed to generate 143 L of H₂ gas, we use the ideal gas law equation to calculate the moles of H₂ gas and then convert to grams using the molar mass of CaH₂. The answer is approximately 271 g.
Step-by-step explanation:
To calculate the grams of CaH₂ needed to generate 143 L of H₂ gas, we first need to use the ideal gas law equation PV = nRT to find the moles of H₂ gas.
Rearranging the equation, n = PV / RT. We are given the pressure of H₂ gas as 110 kPa, the volume as 143 L, and the temperature as 22 °C.
We need to convert the temperature to Kelvin, so 22 °C + 273.15 = 295.15 K.
Next, we can substitute the values into the equation:
n = (110 kPa)(143 L) / ((8.31 J/(mol K))(295.15 K))
Simplifying, we calculate the moles of H₂ gas as approximately 6.42 mol.
Finally, to find the grams of CaH₂ needed, we use the balanced chemical equation 1 mol CaH₂ = 1 mol H₂.
The molar mass of CaH₂ is approximately 42.1 g/mol.
Therefore, the grams of CaH₂ needed is approximately 6.42 mol * 42.1 g/mol, which is equal to approximately 270.582 g.
Rounded to three significant figures, the answer is 271 g.