[hoá 8]Cân bằng số nguyên tử khó!!

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whitetigerbaekho

1, Fex0y + yC0=xFe+y C02
2, 2M + 2xHCl -> 2MClx +H2
3, (6x-4y) Al + 3x Fe2O3 --> (3x-2y) Al2O3 + 6 FexOy
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H

habangbn

1. x=3,y=4 $Fe_3O_4$ + $4CO$ -> $3Fe$ + $4CO_2$
hoặc x=2,y=3 $Fe_2O_3$ + $3CO$ -> $2Fe$ + $3CO_2$
2.x=2 M+ $2HCl$ -> $MCl_2$ + $H_2$
3.x=2,y=3 $2Fe_2O_3$ + 2Al -> $Fe_2O_3$ + $Al_2O_3$
Mình k biết đúng hay sai,nếu đúng nhớ thanks nhá
 
M

my_nguyen070

Hoa

3,
$xFe_2O_3+$\ frac{2x-4y}{3}$Al-----------> 2FexOy+ \frac{x- 2y}{3}Al_2O_3$
 
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T

thong147

1, Fex0y + yCO -> xFe + yCO2
2, 2M + 2xHCl -> 2MClx +xH2
3, xFe2O3+2x-4y/3Al−−−−−−−−−−−>2FexOy+x-2y/3Al2O3
 
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thong006

Fex0y + yCO -> xFe + yCO2
2, 2M + 2xHCl -> 2MClx +xH2
3, xFe2O3+2x-4y/3Al−−−−−−−−−−−>2FexOy+x-2y/3Al2O3
 
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