giúp em với các anh chị ơi

$\rm a)$

$\rm Fe+2HCl \rightarrow FeCl_2+H_2(1)$

$\rm n_{H_2}=$ $\rm\dfrac{1,344}{22,4}=0,06(mol)$ 

Theo $\rm PTHH(1):$ $\rm n_{Fe}=n_{H_2}=0,06(mol)$

$\rm\rightarrow m_{Fe}=3,36(g)$

$\rm\rightarrow \%Fe=$ $\rm\dfrac{3,36}{13}×100\% \approx 25,85\%$ 

$\rm\rightarrow \%Cu=100\%-25,85\%=74,15\%$

$\rm b)$ 

Theo $\rm PTHH(1):$ $\rm n_{HCl}=2n_{H_2}=2.0,06=0,12(mol)$

$\rm\rightarrow m_{\text{DD HCl}}=$ $\rm\dfrac{0,12.36,5.100\%}{15\%}=29,2(g)$ 

$\rm c)$ $\rm V_{\text{DD HCl}}=$ $\rm\dfrac{29,2}{1,025} \approx 28,49(ml)$ 

Tham khảo :

$Fe+2HCl \to FeCl_2+H_2$

$a,$

$n_{H_2}=\dfrac{1,344}{22,4}=0,06(mol)$

$n_{Fe}=n_{H_2}=0,06(mol)$

`=>`$m_{Fe}=nM=0,06.56=3,36(gam)$

`=>`$\%m_{Fe}=\dfrac{3,36}{13}.100\%≈25,85\%$

`=>`$\%m_{Cu}=100\%-25,85\%=74,15\%$

$b,$

$n_{HCl}=2n_{H_2}=2.0,06=0,12(mol)$

$m_{HCl}=0,12.36,5=4,38(gam)$

`=>`$m_{dd,HCl}=\dfrac{m_{HCl}.100}{C\%}=\dfrac{4,38.100}{15}=29,2(gam)$

$c,$

$V_{dd}=\dfrac{m_{dd,HCl}}{D_{dd}}=\dfrac{29,2}{1,025}≈28,49(ml)$