Answer:
139 g of CaH₂ were needed in the reaction
Explanation:
Determine the reaction:
CaH₂ + 2H₂O → Ca(OH)₂ + 2H₂
1 mol of calcium hidride reacts with 2 moles of water to produce 1 mol of calcium hydroxide and 2 moles of hydrogen
Let's determine the moles of formed hydrogen by the Ideal Gases Law
P . V = n . R . T
P = 823 Torr . 1 atm/760 Torr = 1.08 atm
T = Absolute T° → T°C + 273 → 21°C + 273 = 294K
1.08 atm . 147 L = n . 0.082 . 294K
(1.08 atm . 147 L) / (0.082 . 294K) = n → 6.60 moles
Ratio is 2:1. We make a rule of three:
2 moles of H₂ came from 1 mol of hydride
Then 6.60 moles of H₂ must came from 3.30 moles of hydride (6.60 .1) /2
Let's convert the moles to mass → 3.30 mol . 42.08 g/ 1 mol =
138.8 g ≅ 139 g
