Question

If 254.5 mL of nitrogen gas, measured at 715.2 mmHg and 71.7 C, reacts with excess iodine according to the following reaction, what mass of nitrogen triiodide is produced?
N2(g) + 3I2(s) → 2NI3(s) PICK ONE

A

3.34 g

B

6.68 g

C

32.1 g

D

0.237 g

E

1.67 g

Answer 1

PV = nRT

P = pressure = 715.2 mmHg = 0.941 atm

V = volume = 0.2545 L

n = number of moles

R = Gas constant

T = temperature = 71.7 + 273 = 344.7 K

0.941 * 0.2545 = n * 0.0821 * 344.7

0.239 = n * 28.3

n = 0.239 / 28.3 = 0.00845 mol

From the balanced equation we can say that

1 mole of N2 produces 2 mole of NI3 so

0.00845 mole of N2 will produce

= 0.00845 mole of N2 *(2 mole of NI3 / 1 mole of N2)

= 0.0169 mole of NI3

mass of 1 mole of NI3 = 394.719 g

so the mass of 0.0169 mole of NI3 = 6.68 g

Therefore, the mass of NI3 produced would be 6.68 g

option B is correct