Oblicz masę produktu powstającego w reakcji 40g glinu z tlenem.
mAl=27umAl2O3=102u4Al + 3O2---->2Al2O3108g Al---------204g Al2O340g Al-----------xg Al2O3x = 75,56g Al2O3
pozdrawiam ;)
mAl=27g
mAl2O3=2*27g+3*16g=102g
4Al+3O2-->2Al2O3
4*27g----------2*102g
40g--------------xg
xg=40g*204g/108g
xg=75,6g
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mAl=27u
mAl2O3=102u
4Al + 3O2---->2Al2O3
108g Al---------204g Al2O3
40g Al-----------xg Al2O3
x = 75,56g Al2O3
pozdrawiam ;)
mAl=27g
mAl2O3=2*27g+3*16g=102g
4Al+3O2-->2Al2O3
4*27g----------2*102g
40g--------------xg
xg=40g*204g/108g
xg=75,6g