20.

Đặt n_{CaCO_3}=x(mol);n_{MgCO_3}=y(mol)

->100x+84y=2,84(1)

CaCO_3+2HCl->CaCl_2+CO_2+H_2O

MgCO_3+2HCl->MgCl_2+CO_2+H_2O

Theo PT: n_{CO_2}=x+y={0,7437}/{24,79}=0,03(2)

(1)(2)->x=0,02(mol);y=0,01(mol)

->\%m_{CaCO_3}={0,02.100}/{2,84}.100\%\approx 70,42\%

->\%m_{MgCO_3}=100-70,42=29,58\%

21.

a)

Đặt n_{Fe_2O_3}=x(mol);n_{Fe_3O_4}=y(mol)

->160x+232y=27,6(1)

Fe_2O_3+3CO  $$\xrightarrow{t^o}$$  2Fe+3CO_2

Fe_3O_4+4CO  $$\xrightarrow{t^o}$$  3Fe+4CO_2

Theo PT: n_{CO}=3x+4y={12,395}/{24,79}=0,5(2)

(1)(2)->x=0,1(mol);y=0,05(mol)

->\%m_{Fe_2O_3}={0,1.160}/{27,6}.100\%\approx 57,97\%

->\%m_{Fe_3O_4}=100-57,97=42,03\%

b)

Theo PT: n_{Fe}=2x+3y=0,35(mol)

->m_{Fe}=0,35.56=19,6(g)

22.

a)

Đặt n_{Mg}=x(mol);n_{Fe}=y(mol)

->24x+56y=5,2(1)

Mg+2HCl->MgCl_2+H_2

Fe+2HCl->FeCl_2+H_2

Theo PT: n_{H_2}=x+y={3,7185}/{24,79}=0,15(2)

(1)(2)->x=0,1(mol);y=0,05(mol)

->\%m_{Mg}={0,1.24}/{5,2}.100\%\approx 46,15\%

->\%m_{Fe}=100-46,15=53,85\%

b)

Theo PT: n_{HCl}=2n_{H_2}=0,3(mol)

->V_{dd\ HCl}={0,3}/1=0,3(l)

23.

a)

Đặt n_{Al}=x(mol);n_{Fe}=y(mol)

->27x+56y=5,5(1)

2Al+6HCl->2AlCl_3+3H_2

Fe+2HCl->FeCl_2+H_2

Theo PT: n_{H_2}=1,5x+y={4,958}/{24,79}=0,2(2)

(1)(2)->x=0,1(mol);y=0,05(mol)

->\%m_{Al}={0,1.27}/{5,5}.100\%\approx 49,09\%

->\%m_{Fe}=100-49,09=50,91\%

b)

Theo PT: 

n_{HCl}=2n_{H_2}=0,4(mol)

n_{AlCl_3}=x=0,1(mol)

n_{FeCl_2}=y=0,05(mol)

->m_{dd\ HCl}={0,4.36,5}/{14,6\%}=100(g)

->C\%_{AlCl_3}={0,1.133,5}/{5,5+100-0,2.2}.100\%\approx 12,7\%

        C\%_{FeCl_2}={0,05.127}/{5,5+100-0,2.2}.100\%\approx 6,04\%

24.

n_{Na}={2,3}/{23}=0,1(mol)

2Na+2H_2O->2NaOH+H_2

0,1->0,1->0,1->0,05(mol)

->m_{dd\ spu}=2,3+47,8.1-0,05.2=50(g)

->C\%_{NaOH}={0,1.40}/{50}.100\%=8\%