Question

problema 11 va rooooooogggg mult

Answer 1

d=masa molara/28.9 => masa molara = 2.49*28.9 = 72 g/mol
44 g CO2........12 g C
0.305..................x
x = 0,0831 g C
18 g H2O...........2 g H
0,15 g H2O..........y
y = 0,01666 g H
0,0831/12 = 0,006925
0,01666/1 = 0,1666
=>(CH2,4)n
12n+2,4n = 72 => n = 5
FM = C5H12

44 g dioxid.......12 g C
0.44 g..............x
x = 0,12 g C
18 g apa..........2 g H
0.45 g...............y
y = 0,05 g H
n=[0,11*(850/760)]/(0.082*300) = 0,005 moli
m N = 0,005 * 28 = 0.14 g
0,31-0,12-0,14-0,05 = 0 => nu exista oxigen
0,12/12 = 0,01
0,05/1 = 0,05
0,14/14 = 0,01
=> (CH5N)n
Pt n=1 => FM: CH5N