Oblicz ile chlorkó żelaza III powstanie w reakcji 1 mola chloru z zelażem
M(FeCl3)=56g+3*35,5g=162,5g
2Fe+3Cl2-->2FeCl3
1mol-----------------xg
3mol----------------2*162,5g
xg=1mol*325g/3mol
xg=108,3g
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M(FeCl3)=56g+3*35,5g=162,5g
2Fe+3Cl2-->2FeCl3
1mol-----------------xg
3mol----------------2*162,5g
xg=1mol*325g/3mol
xg=108,3g