Đáp án :

Bạn tham khảo nhé

a)`M_(Al_2S_3) = 27.2+32.3=54+96=150 ` g/mol

`%m_(Al) = (27.2.100%)/150 =36%`

`%m_S = 150% - 36% = 114%`

b) ` M_(MgCO_3) = 24+12+16*3=84` g/mol

`%m_(Mg) = (24*100%)/84 approx 28,6%`

`%m_(C) = (12*100%)/84 approx 14,3%`

`%m_(O) = 100% - 28,6% - 14,3% = 57,1%`

c) `M_(AgNO_3) = 108 + 14 + 16.3 = 170` g/mol

`%m_(Ag) = (108*100%)/170 approx 63,5%`

`%m_(N) = (14 * 100%)/170 approx 8,2%`

`%m_(O) = 100% - 63,5% - 8,2% = 28,3%`

d) `M_(Na_2CO_3) = 23.2 + 12 + 16.3 = 106` g/mol

`%m_(Na) = (23.2.100%)/106 = 43,4%`

`%m_(C) = (12*100%)/106 approx 11,3%`

`%m_(O) = 100% - 43,4% - 11,3% = 45,3%`

e) `M_(Zn(NO_3)_2) = 65+(14+16.3).2 = 189` g/mol

`%m_(Zn) = (65*100%)/89 approx 34,4%`

`%m_(N) = (14*2*100%)/189 = 14,8%`

`%m_O = 100% - 34,4% - 14,8% = 50,8%`

f) `M_(Fe(OH)_3) = 56+(16+1).3 = 107` g/mol

`%m_(Fe) = (56*100%)/107 approx 52,3%`

`%m_(O) = (16.3.100%)/107 approx 44,9%`

`%m_(H) = 100% - 52,3% - 44,9% = 2,8%`

 

Có MAl2S3 = 54+96 = 150 (gam/mol)

=}%Al2= mAl2/mAl2S3 . 100 = 54/150 .100= 36%

=}%S3= 100% - 36% = 64%

Có MMgCO3= 24+12+48= 84 (gam/mol)

=} %Mg= 24/84 .100 = 28,57%

=}%C= 12/84 .100 = 14,28%

=}%O3= 100% - (28,57%+14,28%)=57,15%

 

Giải thích các bước giải:

 Còn lại bạn làm nốt nhà :))