reacting mass 有問題

首先 no. of mole =massof a substance(g)/molar mass of the substance

Find the no of moles of Al2(CO3)3
no of moles of Al2(CO3)3
=3/(27x2+12x3+16x9)
=3/234
=0.0128 mol

Find the no of moles of Al^3+ ion and CO3^2-
no. of formula units of Al2(CO3)30.0128x(6.02x10^23)
=7.71x10^21
one formula unit of Al2(CO3)3 contains 2 Al^3+ ion
no.of Al^3+ ion
=2x7.71x10^21
=1.54x10^22
no of mol of Al^3+ ion
=1.54x10^22/L
=1.54x10^22/(6.02x10^23)
=2.56 mol

one formula unit of Al2(CO3)3 contains 3 CO3^2-
no.of CO3^2-
=3x7.71x10^21
=2.31x10^22
no of mol of CO3^2-
=2.31x10^22/(6.02x10^23)
=3.84 mol





2010-09-13 17:30:03 補充：
no.of mol of CO3^2-=3.84 mol
no of CO3^2=3.84 x(6.02x10^23)=2.31x10^24
no of C atoms present =3x2.31x10^24=6.93x10^24
no of mol of C atoms=6.93x10^24/(6.02x10^23)=1.15 mol

no of O atoms present =9x2.31x10^24=2.08x10^25
no of mol of O atoms=2.08x10^25/(6.02x10^23)=3.45 mol

2010-09-13 17:30:06 補充：
no.of mol of CO3^2-=3.84 mol
no of CO3^2=3.84 x(6.02x10^23)=2.31x10^24
no of C atoms present =3x2.31x10^24=6.93x10^24
no of mol of C atoms=6.93x10^24/(6.02x10^23)=1.15 mol

no of O atoms present =9x2.31x10^24=2.08x10^25
no of mol of O atoms=2.08x10^25/(6.02x10^23)=3.45 mol