Question

prin reducerea cu hidrogen a 2,24 g de amestec, format din oxid feros si oxid cupric se formeaza 1,76 g de amestec solid.Sa se calculeze compozitia procentuala masica si molara a amesteculuide oxizi.

Answer 1

FeO+H2= Fe+H2O
CuO +H2= Cu+H2O
 MFeO=56+16=72g/mol
MCuO=64+16=80g/mol
72 g FeO........................56gFe
a.....................................x
x=56*a/72
80g CuO...........................64 gCu
b.......................................y
y=64*b/80
a+b=2,24
x+y=1,76    56a/72+64b/80=1,76  56*80*a+64*72*b=1,76*72*80

a+b=2,24
4480a+4608b=10137,6    /:64
70a+72b=158,4     /:10
7a+7,2b=15,84
a+b=2,24/ *-7
-7a-7b=-15,68
7a+7,2b=15,84
/  0,2b=0,16
a=0,16/0,2=0,8 g FeO
B=2,24-0,8=1,44g CuO

2,24 G AMESTEC.................0,8 FeO..............1,44 gCuO
100.........................................p1.......................p2
p1=0,8*100/2,24=35,71%FeO
p2=1,44*100/2,24=64,28%CuO