oblicz masę tlenu zawartego w 55g octanu etylu.
M(CH3COOC2H5)=4*12g+2*16g+8*1g=48g+32g+8g=88g
32gO₂---------------88g
xgO₂----------------55g
xg=32g*55g/88g
xg=20g
masa octanu etylu: 88g/mol (12+1*3+12+16+16+12+2*1+12+3*1)
masa tlenu = 32g/mol
88g ------------- 32g
55g ------------- x
x = 55*32/88
x= 20g
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M(CH3COOC2H5)=4*12g+2*16g+8*1g=48g+32g+8g=88g
32gO₂---------------88g
xgO₂----------------55g
xg=32g*55g/88g
xg=20g
masa octanu etylu: 88g/mol (12+1*3+12+16+16+12+2*1+12+3*1)
masa tlenu = 32g/mol
88g ------------- 32g
55g ------------- x
x = 55*32/88
x= 20g