\(n_{H_2}=\dfrac{11,2}{22,4}=0,5mol\)

Gọi \(\left\{{}\begin{matrix}n_{Al}=x\\n_{Mg}=y\end{matrix}\right.\)

\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

 x                                                 3/2x

\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)

 y                                           y ( mol )

Ta có:

\(\left\{{}\begin{matrix}27x+24y=10,2\\\dfrac{3}{2}x+y=0,5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,2\end{matrix}\right.\)

\(\Rightarrow m_{Al}=0,2.27=5,4g\)

\(\Rightarrow m_{Mg}=0,2.24=4,8g\)

=> Chọn C

\(PTHH:H_2+CuO\underrightarrow{t^o}Cu+H_2O\)

Áp dụng định luật bảo toàn khối lượng, ta có:

\(m_{H_2}+m_{CuO}=m_{Cu}+m_{H_2O}\\ m_{H_2O}=m_{H_2}+m_{CuO}-m_{Cu}=2+80-64=18\left(g\right)\\ \Rightarrow D\)

\(n_{H_2}=\dfrac{2,24}{22,4}=0,1mol\)

\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

0,1                                         0,1   ( mol )

\(m_{Fe}=0,1.56=5,6g\)

\(\rightarrow m_{Cu}=10-5,6=4,4g\)

--> B

\(n_{H_2}=\frac{5,6}{22,4}=0,25(mol)\\ m_{H_2}=0,25.2=0,5(g)\\ BT H:\\ n_{HCl}=2n_{H_2}=0,25.2=0,5(mol)\\ m_{HCl}=0,5.36,5=18,25(g)\\ BTKL:\\ m_{hh}+m_{HCl}=m_{muối}+m_{H_2}\\ 15+18,25=m_{muối}+0,5\\ \to m_{muối}=32,75(g)\\ \to D\)

\(n_{H_2SO_4}=1\cdot0,15=0,15\left(mol\right)\\ PTHH:Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ \Rightarrow n_{Fe}=n_{H_2SO_4}=0,15\left(mol\right)\\ \Rightarrow m_{Fe}=0,15\cdot56=8,4\left(g\right)\left(B\right)\)

\(n_{H_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)

PT: \(2H_2+O_2\underrightarrow{t^o}2H_2O\)

Theo PT: \(n_{H_2O}=n_{H_2}=0,6\left(mol\right)\Rightarrow m_{H_2O}=0,6.18=10,8\left(g\right)\)

→ Đáp án: B

\(n_{CaO}=\dfrac{5,6}{56}=0.1\left(mol\right)\\ CaO+2HCl->CaCl_2+H_2O\\ =>n_{HCl}=0,2\left(mol\right)\\ m_{HCl}=0,2\cdot36.5=7.3\left(g\right)\\ m_{ddHCl}=\dfrac{7,3\cdot100}{14.6}=50\left(g\right)\)

Chọn đáp án A

\(Fe+2HCl\rightarrow FeCl_2+H_2\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Đặt:\left\{{}\begin{matrix}n_{Fe}=x\left(mol\right)\\n_{Al}=y\left(mol\right)\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}56x+27y=2,78\\x+1,5y=0,07\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=0,04\\y=0,02\end{matrix}\right.\\ n_{FeCl_2}=n_{Fe}=0,04\left(mol\right)\\ \Rightarrow m_{FeCl_2}=0,04.127=5,08\left(g\right)\)