\(m_{X.pư}=80.14\%=11,2g\\ n_{H_2}=\dfrac{4,48}{22,4}=0,2mol\\ X+2HCl\xrightarrow[]{}XCl_2+H_2\\ n_X=n_{H_2}=0,2mol\\ M_x=\dfrac{11,2}{0,2}=56\)

X là Fe

cho nước dư vào hỗn hợp, quấy đều, lọc chất rắn không tan là cát. cô cặn dung dịch thu được tinh thể trắng là muối

\(n_{H_2}=\dfrac{7,437}{24,79}=0,3mol\\ 2R+6HCl\xrightarrow[]{}2RCl_3+3H_2\\ n_R=\dfrac{2}{3}n_{H_2}=0,2mol\\ M_R=\dfrac{5,4}{0,2}=27\Rightarrow R.là.Al\)

a.3 ko pư

b.1,2 thiếu đk

c.1,2,5 thiếu đk

\(a.m_C=\dfrac{8,8}{44}.12=2,4g\\ m_H=\dfrac{3,6}{18}.2=0,4g\\ m_C+m_H=2,4+0,4=2,8g< m_A\)

Trong A có C,H,O

\(b.m_O=4,4-2,8=1,6g\\ M_A=\dfrac{6,6}{1,68:22,4}=88g/mol\\ CTPT\left(A\right):C_xH_yO_z\\ \dfrac{12x}{2,4}=\dfrac{y}{0,4}=\dfrac{16z}{1,6}=\dfrac{88}{4,4}\\ \Rightarrow x=4;y=8;z=2\\ CTPT\left(A\right):C_4H_8O_2\)

\(n_{Mg}=a;n_{Fe}=b\\ n_{H_2}=\dfrac{12,395}{24,79}=0,5mol\\ Mg+2HCl\rightarrow2MgCl_2+H_2\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ \rightarrow\left\{{}\begin{matrix}a+b=0,5\\24a+56b=20\end{matrix}\right.\\ \Rightarrow a=b=0,25mol\\ \%m_{Mg}=\dfrac{0,25.24}{20}\cdot100\%=30\%\\ =>\%m_{Fe}=70\%\)

\(a.n_{Al.pư}=a\\ 2Al+3CuSO_4\rightarrow Al_2\left(SO_4\right)_3+3Cu\\ \Rightarrow m_{\Delta}=\dfrac{3}{2}\cdot64a-27a=46,38-45\\ \rightarrow a=0,02mol\\ n_{Cu}=n_{CuSO_4}=\dfrac{3}{2}n_{Al}=0,03mol\\ m_{Cu}=0,03.64=1,92g\\ b.C_{M_{CuSO_4}}=\dfrac{0,03}{0,4}=0,075M\)

\(n_{AgNO_3}=0,1.0,2=0,02mol\\ AgNO_3+KBr\xrightarrow[]{}AgBr+KNO_3\\ =>\dfrac{0,02}{1}< \dfrac{0,3}{1}=>KBr.dư\\ n_{AgBr}=n_{AgNO_3}=0,02mol\\ m_{AgBr}=0,02.188=3,76g\)