Đáp án + Giải thích các bước giải:

$a\big)Fe+2HCl\to FeCl_2+H_2$

$b\big)$

$n_{Fe}=\frac{5,6}{56}=0,1(mol);n_{HCl}=\frac{5,475}{36,5}=0,15(mol)$

Vì $n_{Fe}>n_{HCl}:2\to Fe$ dư

Theo PT: $n_{Fe(dư)}=0,1-0,15:2=0,025(mol)$

$\to m_{Fe(dư)}=0,025.56=1,4(g)$

$c\big)$

Theo PT: $n_{H_2}=n_{HCl}:2=0,075(mol)$

$n_{PbO}=\frac{22,3}{223}=0,1(mol)$

$PbO+H_2\xrightarrow{t^o}Pb+H_2O$

Vì $n_{PbO}>n_{H_2}\to PbO$ dư

Theo PT: $n_{Pb}=0,075(mol);n_{PbO(dư)}=0,1-0,075=0,025(mol)$

$\to m_{CT}=0,075.207+0,025.223=20,654(g)$

Đáp án+ Giải thích các bước giải:

$a)$ 

$Fe + 2HCl$ $\rightarrow$ $FeCl_2 + H_2$

$b)$

$n_{Fe}$= $\frac{5,6}{56}$= $0,1(mol)$

$n_{HCl}$= $\frac{5,475}{36,5}$= $0,15(mol)$

Tỉ lệ: $\frac{0,1}{1}$ > $\frac{0,15}{2}$   $\rightarrow$ $Fe$ dư.

$n_{Fe(dư)}$= $0,1-0,075= 0,025(mol)$

$\rightarrow$ $m_{Fe(dư)}$= $0,025.56= 1,4(g)$

$c)$

$PbO + H_2$ $\xrightarrow{t^o}$ $Pb + H_2O$

$n_{PbO}$= $\frac{22,3}{223}$= $0,1(mol)$

Tỉ lệ: $\frac{0,1}{1}$ > $\frac{0,075}{1}$   $\rightarrow$ $PbO$ dư.

Theo pt: $n_{Pb}$=  $0,075(mol)$

$n_{PbO(dư)}$= $0,1-0,075=0,025(mol)$

$\rightarrow$ $m_{cr}$= $0,075.207+0,025.223=20,654(g)$