If boron hydride B4H10 is treated with pure oxygen it burns to give B2O3 and H2O.Please help final exam

Question

If a 0.050 g sample of the boron hydride burns completely in O2 what will be the pressure of the gaseous water in 4.25 L flask at 30 degree centigrade? 2B4H10(s) 11O2(g)-->4B2O3 10H2O(g)

Answer 1

2B4H10  +  11O2  ==>  4B2O3  +  10H2O

0.005g B4H10 x (1 mole B4H10 / 53.3234g) x (10 moles H2O / 2 moles B4H10) = 9.38E-5 moles H2O

PV = nRT

P = nRT / V

P = 9.38E-5 moles x 0.0821 L atm/mol K x 303K / 4.25L

P = 5.5 x 10^-4 atm

Answer 2

First thing you  must do, is change to the correct units.

pV=nRT

p=Pa     V=m^3     n=mol    R=Gas constant=8.314J/k/mol     T=K

pV=nRT

using stoichiometry, work out mol of B4H10, then the mol of H20

2.3x10^-3mol

p=(2.3x10^-3mol x 8.314J/K/mol x (273K+30)) / (4.25x10^-3m^-3)

p=1,363Pa

  =1.36kPa