a)Mg+2HCl→MgCl2+H2 (1)

 Theo đề:$n_{Mg}$=$\frac{m_{Mg}}{M_{Mg}}$=$\frac{4,8}{24}$=0,2(mol)

Theo PTHH(1):$n_{MgCl2}$=$n_{Mg}$=0,2(mol) 

$m_{MgCl2}$=$n_{MgCl2}$×$M_{MgCl2}$=0,2×95=19(g)

c)Theo PTHH(1):$n_{HCl}$=2×$n_{Mg}$=2×0,2=0,4(mol)

$m_{HCl}$=$n_{HCl}$×$M_{HCl}$=0,4×36,5=14,6(g)

d)

Fe2O3+3H2→2Fe+3H2O (2)

Theo PTHH(1):$n_{H2}$=$n_{Mg}$=0,2(mol)

Theo PTHH(2):$n_{Fe}$=$\frac{2}{3}$×$n_{H2}$=$\frac{2}{3}$×0,2≈0,1(mol)

$m_{Fe}$=$n_{Fe}$×$M_{Fe}$=0,1×56=5,6(g)

Bạn tham khảo nha!

a. `Mg + 2HCl -> MgCl_2 + H_2` 

b. $n_{Mg}$= $\dfrac{4,8}{24}$ = $0,2$ $mol$ (1)

$m_{MgCl2}$ = $0,2 × 95$ = $19$ $g$

c. $m_{HCl}$ = $0,4 × 36,5$ = $14,6$ $g$

d. `Fe_2O_3 + 3H_2 -> 2Fe + 3H_2O`

Theo phương trình (1), ta có:

$n_{H_2}$ = $n_{Mg}$ = $0,2$ $mol$

-> $m_{Fe}$ = $0,1 × 56$ = $5,6$ $g$