Calcium hydride reacts with water to form hydrogen gas according to the unbalanced equation below: CaH2(s) + H2O(l) --> Ca(OH)2(aq) + H2(g) This reaction is sometimes used to inflate life rafts, weather balloons, and the like, where a simple, compact means of generating H2 is desired. How many grams of calcium hydride are needed to generate 15.0 L of hydrogen gas at 25 degrees C and 825 torr of pressure?

28 grams CaH₂(s) is required for production of 15L H₂(g) at 25°C and 825Torr.

Explanation:

CaH₂(s) + H₂O(l) => Ca(OH)₂(s) + H₂(g)

Using ideal gas law, PV = nRT

=> moles H₂(g) = PV/RT = [(825/760)Atm](15L)/(0.08206L·Atm·mol⁻¹·K⁻¹)(298K) = 0.6659 mol H₂(g)

From stoichiometry of given equation,        

=> 0.6659 mol H₂(g) requires 0.6659 mole CaH₂(s)

Converting moles to grams, multiply by formula weight,

=> 0.6659 mole CaH₂(s) = 0.6659 mole CaH₂(s) x 42g/mole = 27.966 grams CaH₂(s) ≅ 28 grams CaH₂(s)  (2 sig. figs.)

h let me think about that

i'm assuming that the correct answer is indeed the first one/ "a"..