Question

Aflați raportul atomic, raportul de masa si compozitia procentuala la:
-carbonat de potasiu;
-sulfat de aluminiu;
-iodura de amoniu;
-oxid feric;

Answer 1

K2CO3

raport atomic (r.a.) --->2K:1C:3O=2:1:3

raport de masa (r. m.)----->2AK:AC:3AO=2.39:12:3.16=78:12:48=13:2:8 (se
                                                                                                  simplifica cu 6)

Se calc. masa moleculara MK2CO3=2.39+12+3.16=138

Daca in 138 g K2CO3 se gasesc  78 g K,12 g C,48 g O,atunci in 100 g K2CO3 se gasesc x,y,z .

138g K2CO3------78 g K--------12 g C-------48 g O
100 g K2CO3-------x----------------y-------------z
x=100.78:138=56,52 la suta K
Y=100.12:138=8,69  la suta C
z=100.48:138=34,78 la suta O




Al2(SO4)3

r.a.--->2Al:3S:12O=2:3:12
r.m.--->2AAl :3AS:12AO=2.27:3.32:12.16=54:96:192=9:16:32
                                                                       (se simplifica cu 6)

MAl2(SO4)3 =2.27+3.32+12.16=54+96+192=342
342 gAl2(SO4)3------54 g Al-------96 gS-------192 g O
100 gAl2( SO4)3-------x------------y----------------z
x=100.54:342=15,78 la suta Al
y=100.96:342=28,07 la suta S
Z=100.192:342=56,14 la suta O




NH4 I                                                 I=IOD  are masa atomica 127
r.a.--->1N:4H:1 I =1:4:1
r.m.--->AN:4AH:A I =14:4:127

MNH4 I=14+4 +127=145

145 g NH4 I -------14 gN-------4gH--------127 g I
100g NH4 I----------X ------------y-------------z
x=100.14:145=9,65 la suta N
y=100.4 :145=2,75 la suta H
z=100.127:145=87,58 la suta I        I=IOD



Fe2O3
r.a.--->2Fe:3O=2:3
r.m.--->2AFe:3AO=2.56:3.16=112:48=14:6   ( se simplifica cu 8)

MFe2O3=2.56+3.16=112+48=160

160 gFe2O3-------112 gFe-------48 g O
100 gFe2O3--------x-----------------y
x=100.112:160=70 la suta Fe
y=100.48:160=30la suta O