Đáp án:

 `↓`

Giải thích các bước giải:

 Câu 6 :

`n_{H_2}=(6,72)/(22,4)=0,3(mol)`

Gọi `n_{Mg}=a;n_{Zn}=b`

Ta có : `24a+65b=11,3(1)`

`a.`

PTHH:

`Mg+2CH_3COOH->Mg(CH_3COO)_2+H_2`

`a`                                                                  `->a(mol)`

`Zn+2CH_3COOH->Zn(CH_3COO)_2+H_2`

`b`                                                                 `->b(mol)`

Từ PTHH: `a+b=0,3(2)`

từ (1) và (2) ta suy ra :

`a=0,2(mol)`

`b=0,1(mol)`

`b,`

`\%m_{Mg}=(0,2.24)/(11,3).100\%≈42,48\%`

`\%m_{Zn}=(100-42,48)\%=57,52\%`

`c,`

Từ PTHH :

`n_{CH_3COOH}=2a+2b=2.0,2+2.0,1=0,6(mol)`

`C\%_{CH_3COOH}=(0,6.60)/(200).100\%=18\%`

`n_{Mg(CH_3COO)_2}=a=0,2(mol)`

`n_{Zn(CH_3COO)_2}=b=0,1(mol)`

$m_{\text{dd sau pứ}}=m_A+m_{dd CH_3COOH}-m_{H_2}=11,3+200-0,3.2=210,7(g)$

`\%m_{Mg(CH_3COO)_2}=(0,2.142)/(210,7).100\%≈13,48\%`

`\%m_{Zn(CH_3COO)_2}=(0,1.183)/(210,7).100\%≈8,69\%`

Đáp án + Giải thích các bước giải:

$a\big)$

$Mg+2CH_3COOH\to (CH_3COO)_2Mg+H_2$

$Zn+2CH_3COOH\to (CH_3COO)_2Zn+H_2$

$b\big)$

Đặt $n_{Mg}=x(mol);n_{Zn}=y(mol)$

$\to 24x+65y=11,3(1)$

Theo PT: $x+y=n_{H_2}=\frac{6,72}{22,4}=0,3(2)$

Từ $(1)(2)\to x=0,2(mol);y=0,1(mol)$

$\to\%m_{Mg}=\frac{0,2.24}{11,3}.100\%\approx 42,48\%$

$\to \%m_{Zn}=100-42,48=57,52\%$

$c\big)$

Theo PT: `\sum n_{CH_3COOH}=2\sum n_{H_2}=0,6(mol)`

$\to C\%_{CH_3COOH}=\frac{0,6.60}{200}.100\%=18\%$

Theo PT: $n_{(CH_3COO)_2Mg}=x=0,2(mol);n_{(CH_3COO)_2Zn}=y=0,1(mol)$

$m_{dd\,spu}=11,3+200-0,3.2=210,7(g)$

$\to C\%_{(CH_3COO)_2Mg}=\frac{0,2.142}{210,7}.100\%\approx 13,48\%$

$\to C\%_{(CH_3COO)_2Zn}=\frac{0,1.183}{210,7}.100\%\approx 8,69\%$