Question

How many grams of calcium hydride are required to produce 4.56 L of hydrogen gas at 25.0°C and 0.975 atm pressure according to the chemical equation shown below?

CaH₂(s) + 2 H₂O(l) → Ca(OH)₂(aq) + 2 H₂(g)

Answer 1

To produce 4.56 L of hydrogen gas, 20.72 grams of calcium hydride (CaH₂) are required.

Step-by-step explanation:

To answer this question, we need to use the stoichiometry of the chemical equation. From the balanced equation, we can see that 1 mole of CaH₂ reacts to produce 2 moles of H₂ gas. We can use this information to convert the given volume of H₂ gas to moles. Using the ideal gas law equation PV = nRT, we can calculate the number of moles of H₂ gas. Finally, we can use the molar mass of CaH₂ to convert moles to grams.

The molar mass of CaH₂ is 42.1 g/mol. Given that 1 mole of CaH₂ reacts to produce 2 moles of H₂ gas, we can use the stoichiometry to determine that 1 mole of H₂ gas is equal to 21.01 g. So, to find the mass of H₂ gas produced from 4.56 L of H₂ gas, we can set up the following conversion:

• 4.56 L H₂ gas × (1 mole H₂ gas / 22.4 L) × (21.01 g H₂ gas / 1 mole H₂ gas) = 20.72 g H₂ gas

Therefore, 20.72 grams of calcium hydride (CaH₂) are required to produce 4.56 L of hydrogen gas according to the given chemical equation.