Giải hộ mình vớiiiii

6.

Đặt `n_{Na_2CO_3}=x(mol);n_{NaHCO_3}=y(mol)`

`->106x+84y=19(1)`

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

`NaHCO_3+HCl->NaCl+CO_2+H_2O`

Theo PT: `n_{CO_2}=x+y={4,92.1}/{0,082(273+27)}=0,2(2)`

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

`->m_{Na_2CO_3}=0,1.106=10,6(g);m_{NaHCO_3}=0,1.84=8,4(g)`

Theo PT: `n_{HCl\ pu}=2x+y=0,3(mol)`

`->n_{HCl\ du}=0,3.10\%=0,03(mol)`

`->V_{dd\ HCl\ bd}={0,3+0,03}/1=0,33(L)=330(mL)`

7.

Đặt `n_{Mg}=x(mol);n_{Zn}=y(mol)`

`->24x+65y=11,3(1)`

`Mg+H_2SO_4->MgSO_4+H_2`

`Zn+H_2SO_4->ZnSO_4+H_2`

Theo PT: `n_{H_2}=x+y={7,38.1}/{0,082(273+27)}=0,3(2)`

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

`->m_{Mg}=0,2.24=4,8(g);m_{Zn}=0,1.65=6,5(g)`

Theo PT: `n_{H_2SO_4\ pu}=n_{H_2}=0,3(mol)`

`->n_{H_2SO_4\ du}=0,3.10\%=0,03(mol)`

`->V_{dd\ H_2SO_4\ bd}={0,03+0,3}/1=0,33(L)=300(mL)`

$6.$

`-` $HCl$ dư nên không chia trường hợp.

$NaHCO_3:a(mol);Na_2CO_3:b(mol)$

`->`$84a+106b=19$$(1)$

$pV=nRT$

`->`$n_{CO_2}=\dfrac{4.92 \times 1}{0.0821 \times (27+273)}=0.2(mol)$

$NaHCO_3+HCl \to H_2O+CO_2+NaCl$

$Na_2CO_3+2HCl \to CO_2+H_2O+2NaCl$

`->`$a+b=0.2$$(2)$

`-` Từ $(1),(2)$ suy ra: $a=b=0.1(mol)$

`=>`$m_{NaHCO_3}=8.4(gam);m_{Na_2CO_3}=10.6(gam)$

$n_{HCl,pu}=0.1 +0,1 \times 2=0.3(mol)$

`=>`$n_{HCl,đầu}=0.3 \times 110\%=0.33(mol)$

`=>`$V_{HCl,đầu}=0.33(L)=330(mL)$

$2.$

$pV=nRT$

`->`$n_{H_2}=\dfrac{7.38 \times 1}{0.0821 \times (27+273)}=0.3(mol)$

$Mg:a(mol);Zn:b(mol)$

`->`$24a+65=11.3$$(1)$

$Mg+H_2SO_4 \to MgCl_2+H_2$

$Zn+H_2SO_4 \to ZnCl_2+H_2$

`->`$a+b=0.3$$(2)$

`-` Từ $(1),(2)$ suy ra: $a=0.2;b=0.1$

`=>`$m_{Mg}=4.8(gam);m_{Zn}=6.5(gam)$

$n_{H_2SO_4,pu}=a+b=0.3(mol)$

`->`$n_{H_2SO_4,đầu}=0.3 \times 110\%=0.33(mol)$

`=>`$V_{H_2SO_4,đầu}=0.33(L)=330(mL)$