Question

How many grams ofNH

3

N

H

3

 can be produced when 3.60 L

L

 of NO

2

N

O

2

 reacts at 415 

∘

C

∘

C

 and 695 mmHg

m

m

H

g

 ?

Answer 1

2NO₂ + 7H₂ → 2NH₃ + 4H₂O

$V= 3.60 \;L \\ T=415 + 273 = 688 \;K \\ p=695 \;mmHg = 0.914 \;atm$

Ideal gas law

pV =nRT

R = 0.08206 L×atm/mol×K

$n = \frac{pV}{RT} \\ n(NO_2) = \frac{0.914 \times 3.60}{0.08206 \times 688} = 0.0583 \;mol$

According to the reaction:

n(NH₃) = n(NO₂) = 0.0583 mol

MM(NH₃) = 17.0 g/mol

m(NH₃) $= 0.0583 \times 17.0 = 0.990 \; g ≈ 1 \;g$

Answer: 1 g