Răspuns:
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A.Fe= 55g/mol
A,S=32g/mol
miu,FeS= 87g/mol
-calculez masa de Fe pur
p=m,px100/m,i---> m,p= 175x80/100g=140g
-calculez masa de Fe si masa de S care intra in reactia din care se obtin 176g FeS
55g.......32g..........87g
Fe + S-------> FeS
x.................y..............176g
x= 111,3g Fe
y= 64,7g S
-deci, intra in reactie 64,7gS si raman (140-111,3)g= 28,7g Fe
m,ametsec final= 28,7gFe+176gFeS= 204,7g
-calculez compozitia procentuala a amestecului final
in 204,7g ....... 28,7gFe.............176gFeS
100g.......................x..............................y
x= 100x28,7/204,7g=....
y= 100x176/204,7g=....
finalizeaza calculele !!!!!!
100g.........................x.......................y
Explicație:
