Help me please... =[?

Question

Calcium hydride, CaH2, reacts with water to form hydrogen gas.
CaH2(s) + 2 H2O(l) Ca(OH)2(aq) + 2 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 CaH2 are needed to generate 17.0 L of H2 gas if the pressure of H2 is 760. torr at 28°C?
______ g

Answer 1

First, find the moles of H2 using the ideal gas law (760 torr = 1 atm):

n=PV/RT = (1 atm)(17.0L)/(0.0821 L*atm/mol*K)(301.15K)
n of H2 = 0.6876 mol H2

Then stoichiometry:

(0.6876 mol H2)(1 mol CaH2/2 mol H2)(42.094 g CaH2/1 mol CaH2) = 14.5g CaH2 needed.

Answer 2

Converting to STP with P constant,
V1/V2 = T1/T2
V2 = 17*301/293 = 17.464 L
x/17.464 L = 2.016 g/22.4L
x = 2.016*17.464 g/22.4 = 1.5718 g H2
Now
x/(40.08+2*1.008) = 1.5718/4.032
x = 42.096* 1.5718/4.032
x = 16.410 g CaH2