ILE TLENU ZUŻYTO DO CAŁKOWITEGO SPALENIA 23 G ETANOLU?
mC2H5OH=2*12u+6*1u+16u=46u
mO2=32u
C2H5OH+3O2-->2CO2+3H2O
46u------3*32u
23g-------xg
xg=23u*62g/46u
xg=48g
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mC2H5OH=2*12u+6*1u+16u=46u
mO2=32u
C2H5OH+3O2-->2CO2+3H2O
46u------3*32u
23g-------xg
xg=23u*62g/46u
xg=48g