Thi???

ta có :

m _Cl2=41,7-20,4=21,3g

=>n _Cl2=\(\dfrac{21,3}{71}\)=0,3 mol

=>V_Cl2=0,3.22,4=6,72l

\(n_C=n_{CO_2}=\dfrac{4,4}{44}=0,1\left(mol\right)\\ n_H=2n_{H_2O}=2.\dfrac{1,8}{18}=0,1\left(mol\right)\\ Xét:m_C+m_H=0,1.12+0,2=1,4\left(g\right)\)

=> A chỉ có C và H

\(CTPT:C_xH_y\\ \Rightarrow x:y=0,1:0,2=1:2\\ \Rightarrow\left(CH_2\right)_n=14.2=28\left(\dfrac{g}{mol}\right)\\ \Rightarrow n=2\\ CTPT:C_2H_4\)

Đặt công thức tổng quát: C_xH_yO_z ( x,y ∈ N^* , z ∈ N )

Ta có

m_C =\(\dfrac{4,4.3}{11}=1,2g\)

m_H =\(\dfrac{1,8}{9}=0,2g\)

m_C + m_H = 1,2 + 0,2 = 1,4

⇒ hợp chất không có oxi

Đặt tỉ lệ ta có

\(\dfrac{12x}{1,2}=\dfrac{y}{0,2}=\dfrac{28}{1,4}\)

⇒ x = 2 ; y = 4

⇒ CTPT: C₂H₄

a)

\(n_{CH_4}=n_{C_xH_y}=\dfrac{0,5}{22,4}\left(mol\right)\)

=> \(\dfrac{0,5}{22,4}.16+\dfrac{0,5}{22,4}.M_{C_xH_y}=0,9375\)

=> M_CxHy = 26 (g/mol)

=> x = 2, y = 2 thỏa mãn

CTPT: C₂H₂

b) 

\(n_{C_2H_2}=\dfrac{\dfrac{8,96}{2}}{22,4}=0,2\left(mol\right)\)

x = m_tăng = m_C2H2 = 0,2.26 = 5,2 (g)

PTHH: C₂H₂ + 2Br₂ --> C₂H₂Br₄

             0,2----->0,4

=> n_Br2(tt) = \(\dfrac{0,4.120}{100}=0,48\left(mol\right)\)

=> \(V_{dd.Br_2}=\dfrac{0,48}{2}=0,24\left(l\right)\)

c) \(n_{CH_4}=n_{C_2H_2}=0,2\left(mol\right)\)

PTHH: CH₄ + 2O₂ --t^o--> CO₂ + 2H₂O

            0,2--->0,4

=> V_O2 = 0,4.22,4 = 8,96 (l)

=> V_kk = 8,96 : 20% = 44,8 (l)

\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)

a          2a          a            a

\(FeS+2HCl\rightarrow FeCl_2+H_2S\uparrow\)

b            2b            b            b

\(n_{HCl}=\dfrac{400\times7.3}{100\times36.5}=0.8mol\)

\(n_X=\dfrac{4.48}{22.4}=0.2mol\)

\(M_X=2\times9=18\Leftrightarrow\dfrac{2a+34b}{a+b}=18\)

Ta có: \(\left\{{}\begin{matrix}a+b=0.2\\\dfrac{2a+34b}{a+b}=18\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a+b=0.2\\2a+34b=3.6\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a=0.1\\b=0.1\end{matrix}\right.\)

a. \(\%V_{H_2}=\dfrac{0.1\times22.4\times100}{4.48}=50\%\)

\(\%V_{H_2S}=100-50=50\%\)

b. \(a=0.1\times56+0.1\times88=14.4g\)

\(\%m_{Fe}=\dfrac{0.1\times56}{14.4}\times100=38.8\%\)

\(\%m_{FeS}=100-38.8=61.2\%\)

c. m dung dịch sau phản ứng\(=14.4+400-0.1\times2-0.1\times34=410.8g\)

nHCl phản ứng\(=2\times0.1+2\times0.1=0.4mol\)

nHCl dư = 0.8 - 0.4 = 0.4 mol

\(C\%_{HCldu}=\dfrac{0.4\times36.5\times100}{410.8}=3.55\%\)

\(C\%_{FeCl_2}=\dfrac{0.2\times127\times100}{410.8}=6.18\%\)

\(m_{HCl}=14,6\%.300=43,8\left(g\right)\\ n_{HCl}=\dfrac{43,8}{36,5}=1,2\left(mol\right)\\ PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\left(1\right)\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\left(2\right)\\ Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\left(3\right)\\ Theo.pt\left(1,2,3\right)=n_{H_2}=\dfrac{1}{2}n_{HCl}=\dfrac{1}{2}.1,2=0,6\left(mol\right)\\ m_{H_2}=0,6.2=1,2\left(g\right)\\ V_{H_2}=0,6.22,4=13,,44\left(l\right)\)

Áp dụng ĐLBTKL, ta có:

\(m_{kl\left(Mg,Al,Zn\right)}+m_{HCl}=m_{muối\left(MgCl_2,AlCl_3,ZnCl_2\right)}+m_{H_2}\\ \Rightarrow m_{muối}=14,3+43,8-1,2=56,9\left(g\right)\)

\(a,n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\ PTHH:Ca+2H_2O\rightarrow Ca\left(OH\right)_2+H_2\uparrow\left(1\right)\\ Theo.pt\left(1\right):n_{Ca}=n_{H_2}=0,1\left(mol\right)\\ \Rightarrow m_{Ca}=0,1.40=4\left(g\right)\\ \Rightarrow\%m_{Ca}=\dfrac{4}{13,6}=29,41\%\\ \%m_{CaO}=100\%-29,41\%=70,59\%\\ b,Thiếu.dữ.kiện.về.m_{H_2O}\)

\(n_{CO_2}=\dfrac{4,48}{22,4}=0,2mol\)

\(MgCO_3+2HCl\rightarrow MgCl_2+CO_2+H_2O\)

\(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\)

\(\Rightarrow\left\{{}\begin{matrix}84\cdot n_{MgCO_3}+100\cdot n_{CaCO_3}=18,4\\n_{MgCO_3}+n_{CaCO_3}=n_{CO_2}=0,2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}n_{MgCO_3}=0,1mol\\n_{CaCO_3}=0,1mol\end{matrix}\right.\)

\(\%m_{CaCO_3}=\dfrac{0,1\cdot100}{18,4}\cdot100\%=54,35\%\)

\(\%m_{MgCO_3}=100\%-54,35\%=45,65\%\)