a,

$n_{Fe}=\dfrac{1,68}{56}=0,03(mol)$

$n_{Cl_2}=\dfrac{0,84}{22,4}=0,0375(mol)$

$2Fe+3Cl_2\buildrel{{t^o}}\over\to 2FeCl_3$

$\dfrac{0,0375}{3}<\dfrac{0,03}{2}$

$\to Cl_2$ hết, $Fe$ dư 

$n_{Fe\text{pứ}}=n_{FeCl_3}=\dfrac{2}{3}n_{Cl_2}=0,025(mol)$

$\to m_{FeCl_3}=0,025.162,5=4,0625g$

$n_{Fe\text{dư}}=0,03-0,025=0,005(mol)$

$\to m_{Fe\text{dư}}=0,005.56=0,28g$

b,

$Fe+2HCl\to FeCl_2+H_2$

$\to n_{HCl}=2n_{Fe\text{dư}}=0,01(mol)$

$\to V_{dd HCl}=\dfrac{0,01}{0,5}=0,02l=20ml$

Đáp án:

a,

`-` `m_{Fe\ \text{(dư)}}=0,28\ (g).`

`-` `m_{FeCl_3}=4,0625\ (g).`

b, `V_{\text{dd HCl}}=0,02\ (l)=20\ (ml).`

Giải thích các bước giải:

`-` `n_{Fe}=\frac{1,68}{56}=0,03\ (mol).`

`-` n_{Cl_ư=\frac{0,84}{22,4}=0,0375\ (mol).`

a,

Phương trình hóa học:

`2Fe + 3Cl_2 \overset{t^o}\to 2FeCl_3`

`-` Tỉ lệ: `n_{Fe}:n_{Cl_2}=\frac{0,03}{2}>\frac{0,0375}{3}`

`\to Fe` dư, `Cl_2` hết.

`\to n_{Fe\ \text{(dư)}}=0,03-\frac{0,0375\times 2}{3}=0,005\ (mol).`

`\to m_{Fe\ \text{(dư)}}=0,005\times 56=0,28\ (g).`

`-` Theo phương trình: `n_{FeCl_3}=\frac{2}{3}n_{Cl_2}=0,025\ (mol).`

`\to m_{FeCl_3}=0,025\times 162,5=4,0625\ (g).`

b,

Phương trình hóa học:

`Fe + 2HCl \to FeCl_2 + H_2\uparrow`

`-` Theo phương trình: `n_{HCl}=2n_{Fe}=0,01\ (mol).`

`\to V_{\text{dd HCl}}=\frac{0,01}{0,5}=0,02\ (l)=20\ (ml).`

\(\boxed{\text{LOVE TEAM}}\)