$PTHH: Fe+2HCl$`->`$FeCl_{2}+H_{2}$
$n_{Fe}=\frac{m}{M}=\frac{4,2}{56}=0,075 (mol)$
$m_{HCl}=\frac{C\%.mdd}{100\%}=\frac{3,65\%.100}{100\%}=3,65 (g)$
`=>`$n_{HCl}=\frac{m}{M}=\frac{3,65}{1+35,5}=0,1 (mol)$
So sánh tỉ lệ: $\frac{0,075}{1}>\frac{0,1}{2}$`=>`$Fe$ dư
Theo $PTHH$:
$n_{Fe}=\frac{1}{2}n_{HCl}=\frac{1}{2}.0,1=0,05 (mol)$
`=>`$m_{Fe}=n.M=0,05.56=2,8 (g)$
Theo $PTHH$:
$n_{H_{2}}=\frac{1}{2}n_{HCl}=\frac{1}{2}.0,1=0,05 (mol)$
`=>`$m_{H_{2}}=n.M=0,05.2=0,1 (g)$
Áp dụng định luật bảo toàn khối lượng
`=>`$mdd_{FeCl_{2}}=m_{Fe}+m_{HCl}-m_{H_{2}}=2,8+100-0,1=102,7 (g)$
