Đặt $n_{Fe}=x(mol);n_{Zn}=y(mol)$

$\Rightarrow 56x+65y=6,05(1)\\ a,Zn+2HCl\to ZnCl_2+H_2\\ Fe+2HCl\to FeCl_2+H_2\\ \Rightarrow x+y=n_{H_2}=0,1(2)\\ (1)(2)\Rightarrow x=y=0,05(mol)\\ b,\begin{cases} \%_{Fe}=\dfrac{0,05.56}{6,05}.100\%=46,28\%\\ \%_{Zn}=100\%-46,28\%=53,72\% \end{cases}\\ c,n_{FeCl_2}=n_{ZnCl_2}=0,05(mol)(x=y)\\ \Rightarrow m_{FeCl_2}=0,05.127=6,35(g)\\ m_{ZnCl_2}=0,05.136=6,8(g)\\ d,\Sigma n_{HCl}=2x+2y=0,2(mol)\\ \Rightarrow C\%_{HCl}=\dfrac{0,2.36,5}{150}.100\%=4,867\%\\ e,C_{M_{HCl}}=\dfrac{0,2}{0,15}=1,33M$

Fe + 2HCl ra FeCl2 + H2 Cho xy lần lượt là số molFeZn 1. 2. 1. 1 x. x 0.05. 0.1. 0.05 Zn+ 2HCl ra ZnCl2 + H2 1. 2. 1. 1 y. y 0.05 0.1. 0.05 mhh= 6.05g. Suy ra 56x+ 65y= 6.05. (1) x+ y= 0.1(2) Từ 1 và 2 ta có hệ pt 56x+ 65y= 6.05. x+ y= 0.1 Suy ra x= y= 0.05 x = nFe= 0.05mol y= nZn= 0.05mol %mFe= 0.05* 56/6.05*100= 46.28% %mZn= 100-46.28%=53.72% c) m muối tạo thành = mFeCl2 + mZnCl2 = 6.35+6.8= 13.15g D) mctHCl= 0.1 *36.5+0.1*36.5=7.3g C% HCl= 7.3/150*100=4.867% nHCl 2 pt = 0.2mol CmHCl= 0.2/ 0.15=1.33M