Question

If 484.7 mL of nitrogen gas, measured at 777.5 mmHg and 88.2^oC, reacts with excess iodine according to the following reaction, what mass of nitrogen triiodide is produced?

N2(g) + 3I2(s) → 2NI3(s)

Answer 1

substrate -----> product
n 2n

moles of N2 in substrate = 2(moles of NI3 )

now for N2
PV = nRT

P = 777.5/760 = 1.023
V = 0.484 L
R = 0.0821
T = 88.2 + 273 = 361.3

so now moles n = PV/RT
n = 1.023 * 0.484 / 0.0821*361.3
= 0.0167 moles

so no of moles of NI3 = 2*0.0167 = 0.0334

molecular wt of NI3 = 394.7

so mass of NI3 formed = moles * mole. wt
= 0.0334 * 394.7
= 13.18 gm
ANSWER