Question

If 249.8 mL of nitrogen gas, measured at 728.4 mmHg and 19.7oC, reacts with excess iodine according to the following reaction, what mass of nitrogen triiodide is produced? N2(g) + 3I2(s) →2NI3(s)N2(g) + 3I2(s) →2NI3(s)

Select one:

a.
4.77gNI34.77g  NI3

b.
9.87gNI39.87g  NI3

c.
7.86gNI37.86g  NI3

d.
2.33gNI3

Answer 1

PV = nRT

P = 728.4 mmHg = 0.958 atm

V = 249.8 mL = 0.2498 L

T = 19.7 + 273 = 292.7 K

0.958 * 0.2498 = n * 0.0821 * 292.7

0.239 = n * 24.0

n = 0.009958 mole

from the balanced equation we can say that

1 mole of nitrogen gas produces 2 mole of nitrogen triiodide

0.009958 mole of nirogen gas will produce

= 0.009958 mole of nirogen *(2 mole of nitrogen triiodide / 1 mole of nitrogen gas)

= 0.0199 mole of nitrogen triiodide

1 mole of NI3 = 394.719 g

0.0199 mole of NI3 = 7.86 g

Thereforw, the mass of NI3 produced will be 7.86 g

option C is correct