1.

n_{Na_2CO_3}=x(mol);n_{CaCO_3}=y(mol)

->106x+100y=20,6(1)

200cm^3=0,2l

->n_{HCl}=0,2.2=0,4(mol)

Na_2CO_3+2HCl->2NaCl+CO_2+H_2O

CaCO_3+2HCl->CaCl_2+CO_2+H_2O

Theo PT: n_{HCl}=2x+2y=0,4(2)

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

->\%m_{Na_2CO_3}={0,1.106}/{20,6}.100\%\approx 51,46\%

->\%m_{CaCO_3}=100-51,46=48,54\%

2.

Đặt n_{MgO}=x(mol);n_{Zn}=y(mol)

->40x+65y=34(1)

MgO+2HCl->MgCl_2+H_2O

Zn+2HCl->ZnCl_2+H_2

Theo PT: n_{MgCl_2}=n_{Mg}=x(mol);n_{ZnCl_2}=n_{Zn}=y(mol)

->95x+136y=73,4(2)

(1)(2)->x=0,2;y=0,4

->\%m_{MgO}={0,2.40}/{34}.100\%\approx 23,53\%

->\%m_{Zn}=100-23,53=76,47\%

Bài 1:

200cm^3=200ml=0,2l

n_{HCl}=0,2.2=0,4(mol)

PTHH: Na_2CO_3+2HCl->2NaCl+CO_2+H_2O

CaCO_3+2HCl->CaCl_2+H_2O

Gọi x,y là số mol của Na_2CO_3, CaCO_3

Theo bài ra: m_{hh}=106x+100y=20,6(g)(I)

Theo PTHH: n_{HCl}=2n_{Na_2CO_3}+2n_{CaCO_3}

<=>0,4=2x+2y(II)

Từ (I),(II): x=0,1; y=0,1

\%m_{Na_2CO_3}=(0,1.106)/(20,6).100\%=51,46\%

\%m_{CaCO_3}=(0,1.100)/(20,6).100\%=48,54\%

Bài 2:

PTHH:MgO+2HCl->MgCl_2+H_2O

Zn+2HCl->ZnCl_2+H_2

Gọi x,y là số mol của MgO,Zn

m_{hh}=40x+65y=34(g)(I)

Theo PTHH: n_{MgCl_2}=n_{MgO}=x(mol)

n_{ZnCl_2}=n_{Zn}=y(mol)

Theo bài ra: m_{m\uối}=95x+136y=73,4(g)(II)

Từ (I),(II): x=0,2; y=0,4

\%m_{MgO}=(0,2.40)/(34).100\%=23,53\%

\%m_{Zn}=(0,4.65)/(34).100\%=76,47\%