Final answer:
To produce 8.400 grams of hydrogen gas, one would need 87.498 grams of calcium hydride based on the stoichiometric calculations from the balanced reaction equation.
Step-by-step explanation:
The student asked how many grams of calcium hydride (CaH₂) are needed to produce 8.400 grams of hydrogen gas (H₂). To solve this problem, we can use stoichiometry based on the balanced chemical equation:
CaH₂ + 2H₂O → Ca(OH)₂ + 2H₂
From the equation, 1 mole of CaH₂ produces 2 moles of H₂. First, we determine the molar mass of H₂, which is approximately 2.02 g/mol. Then we find the molar mass of CaH₂, which is approximately 42.09 g/mol. Using the given mass of H₂ (8.400 g), we can calculate the moles of H₂ produced:
8.400 g H₂ × (1 mol H₂/2.02 g H₂) = 4.1584 moles H₂
Since 2 moles of H₂ are produced from 1 mole of CaH₂, we need half the amount of moles of CaH₂:
4.1584 moles H₂ × (1 mol CaH₂/2 mol H₂) = 2.0792 moles CaH₂
Now we convert the moles of CaH₂ to grams:
2.0792 moles CaH₂ × (42.09 g CaH₂/1 mol CaH₂) = 87.498 g CaH₂.
Thus, 87.498 grams of calcium hydride are needed to produce 8.400 grams of hydrogen gas.