How many moles of 18l of nh3 at 30 celcius and 912 mmhg? how many grams?

Answer is: mass of ammonia is 14.76 grams and amount is 0.87 moles.
V(NH₃) = 18 L.
T = 30°C = 303.15 K.
p = 912 mmHg ÷ 760 mmHg/atm= 1.2 atm.
R = 0.08206 L·atm/mol·K.
Ideal gas law: p·V = n·R·T.
n = p·V / R·T.
n(NH₃) = 1.2 atm ·18 L / 0.08206 L·atm/mol·K · 303.15 K.
n(NH₃) = 0.87 mol.
m(NH₃) = n(NH₃) · M(NH₃).
m(NH₃) = 0.87 mol · 17 g/mol.
m(NH₃) = 14.76 g.

They are all decomposition reactions.