What is the volume, in mL, of 0.50 moles of ammonia (NH3) at STP?

A. 1120 L

B. 22.32 mL

C. 8500 mL

D. 11,200 mL

For the purpose we will here use the ideal gas law:

p*V=n*R*T

V = ?

n = 0.5 moleT = 273.15 K (at STP)

p = 101.325 kPa (at STP)

R is universal gas constant, and its value is 8.314 J/mol*K

Now when we have all necessary date we can calculate the number of moles:

V=nxRxT/p

V=0.5x8.314x273.15/101.325 = 11.2 L = 11200 mL

Answer: D.