Oblicz, ile gramów bromu może przyłączyć 1,4 g etanu.
mBr₂=2*80g=160g
mC₂H₆=2*12g+5*1g=30g
C₂H₆ + Br₂-->C₂H₅Br + HBr
1,4g----xg
30g-----160g
xg=1,4g*160g/30g
xg=7,5g
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mBr₂=2*80g=160g
mC₂H₆=2*12g+5*1g=30g
C₂H₆ + Br₂-->C₂H₅Br + HBr
1,4g----xg
30g-----160g
xg=1,4g*160g/30g
xg=7,5g