Oblicz ile gram tlenu potrzeba do utlenienia żelaza jeśli powstaje 15g tlenku żelaza (III)
Z góry dzeki :)
mO₂=2*16g=32g
mFe₂O₃=2*56g+3*16g=160g
3O₂+4Fe-->2Fe₂O₃
3*32g-------2*160g
xg-----------15g
xg=96g*15g/320g
xg=4,5g
mO2=32u
mFe2O3=160u
4Fe +3 O2-->2Fe2O3
3*32g tlenu ----- 2*160g Fe2O3
xg tlenu ------- 15g Fe2O3
x = 4,5g Fe2O3
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mO₂=2*16g=32g
mFe₂O₃=2*56g+3*16g=160g
3O₂+4Fe-->2Fe₂O₃
3*32g-------2*160g
xg-----------15g
xg=96g*15g/320g
xg=4,5g
mO2=32u
mFe2O3=160u
4Fe +3 O2-->2Fe2O3
3*32g tlenu ----- 2*160g Fe2O3
xg tlenu ------- 15g Fe2O3
x = 4,5g Fe2O3