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
?
2NO2 + 7H2 → 2NH3 + 4H2O
"V= 3.60 \\;L \\\\\n\nT=415 + 273 = 688 \\;K \\\\\n\np=695 \\;mmHg = 0.914 \\;atm"
Ideal gas law
pV =nRT
R = 0.08206 L×atm/mol×K
"n = \\frac{pV}{RT} \\\\\n\nn(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 \u2248 1 \\;g"
Answer: 1 g
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