\(a,PTHH:2Al+3X_2\rightarrow2AlX_3\\ Theo.ĐLBTKL,ta.có:\\ m_{Al}+m_{X_2}=m_{AlX_3}\\ \Leftrightarrow m_{Al}+33,6=37,38\\ \Leftrightarrow m_{Al}=3,78\left(g\right)\\ \Rightarrow n_{Al}=\dfrac{3,78}{27}=0,14\left(mol\right)\\ n_{X_2}=\dfrac{3}{2}.0,14=0,21\left(mol\right)\\ \Rightarrow M_{X_2}=\dfrac{33,6}{0,21}=160\left(\dfrac{g}{mol}\right)\\ \Rightarrow M_X=80\left(\dfrac{g}{mol}\right)\\ \Rightarrow X:Brom\left(Br=80\right)\\ b,n_{AlBr_3}=\dfrac{42,72}{267}=0,16\left(mol\right)\\ AlBr_3+3AgNO_3\rightarrow Al\left(NO_3\right)_3+3AgBr\downarrow\left(vàng.nhạt\right)\\ n_{AgBr}=n_{AgNO_3}=0,16.3=0,48\left(mol\right)\\ \Rightarrow m_T=m_{\downarrow}=m_{AgBr}=188.0,48=90,24\left(g\right)\)

\(m_{ddZ}=42,72+447,52-90,24=400\left(g\right)\\ n_{Al\left(NO_3\right)_3}=n_{AlBr_3}=0,16\left(mol\right)\\ \Rightarrow C\%_{ddZ}=C\%_{ddAl\left(NO_3\right)_3}=\dfrac{0,16.213}{400}.100=8,52\%\)

\(c,n_{Br_2}=\dfrac{12,8}{160}=0,08\left(mol\right)\\ 2NaI+Br_2\rightarrow2NaBr+I_2\\ n_{I_2}=n_{Br_2}=0,08\left(mol\right);n_{NaI}=2.0,08=0,16\left(mol\right)\\ \Rightarrow x=C_{MddNaI}=\dfrac{0,16}{0,25}=0,64\left(M\right)\\ m_{I_2}=0,08.254=20,32\left(g\right)\)

X + AgNO3\(\rightarrow\)kết tủa AgCl + muối Y

Kết tủa là AgCl\(\rightarrow\) nAgCl=\(\frac{22,96}{\text{108+35,5}}\)=0,16 mol

Bảo toàn Ag: nAgCl=nAgNO3=0,16 mol

\(\rightarrow\)mAgNO3=0,16.(108+62)=27,2 gam

BTKL: mX + mAgNO3=mAgCl + mY

\(\rightarrow\)9,3+27,2=22,96+mY\(\rightarrow\)mY=13,54 gam

a)n_Cl2=0,3mol=>m_Cl2=21,3g

Định luật bảo tòan khối lượng:m_a=m_B-m_Cl2=29,6-21,3=8,3g

b)n_Cl-=2.0,3=0,6mol=>n_AgCl=0,6mol=>m_C=0,6.143,5=86,1g

n_Ag+=n_Cl-=0,6mol=>n_AgNO3=0,6mol=>V_AgNO3=\(\dfrac{0,6}{2}\)=0,3lít

cám ơn nhiều ạ

1/

gọi tên halogen đó là X,ta có:

ZnX\(_2\)+2\(AgNO_3\)→Zn\(\left(NO_3\right)_2\)+2AgX

dựa vào phương trình ta có:

\(\dfrac{32,64}{65+2_{MX}}\)=\(\dfrac{68,88}{2.\left(108+M_x\right)}\)

⇔\(M_x\)=35,5

→clo(Cl)

vậy muối đó là:ZnCl\(_2\)

Câu 1:

Gọi \(\left\{{}\begin{matrix}n_{Cu}:x\left(mol\right)\\n_{Fe}:y\left(mol\right)\end{matrix}\right.\)

\(Cu+Cl_2\rightarrow CuCl_2\)

\(Fe+Cl_2\rightarrow FeCl_2\)

\(\left\{{}\begin{matrix}64x+56y=30,4\\2x+3y=1,2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,3\\y=0,2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Cu}=0,3.64=19,2\left(g\right)\\m_{Fe}=0,2.56=11,2\left(g\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Cu}=63,16\%\\\%m_{Fe}=36,84\%\end{matrix}\right.\)

BTNT Cl:

\(n_{AgCl}=2.n_{Cl2}=1,2\left(mol\right)\)

\(\Rightarrow m_{AgCl}=172,2\left(g\right)\)

Câu 2:

Gọi \(\left\{{}\begin{matrix}n_{Al}:x\left(mol\right)\\n_{Zn}:y\left(mol\right)\end{matrix}\right.\)

\(2Al+6HCl2\rightarrow AlCl_3+3H_2\)

x______________x________3x/2

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

y___________y_________y

\(m_{kl}=27x+65y=3,57\left(1\right)\)

\(m_{muoi}=133,5x+136y=12,09\left(2\right)\)

\(\left(1\right)+\left(2\right)\Rightarrow\left\{{}\begin{matrix}x=0,06\\y=0,03\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Al}=0,06.27=1,62\left(g\right)\\m_{Zn}=0,03.65=1,95\left(g\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%_{Al}=45,38\%\\\%_{Zn}=54,62\%\end{matrix}\right.\)

Bảo toàn e: \(n_{H2}=0,12\left(mol\right)\Rightarrow V=\frac{32}{12}=2,46\left(l\right)\)

11) Theo đề, ta có: \(n_{MnO_2}=\dfrac{78,3}{87}=0,9\left(mol\right)\)

PTHH: \(MnO_2+4HCl\rightarrow MnCl_2+2H_2O+Cl_2\left(1\right)\)

Số mol: 0,9 mol 3,6 mol 0,9 mol 1,8 mol 0,9 mol

a, Theo phương trình, ta có: \(n_{HCl}=4n_{MnO_2}=4.0,9=3,6\left(mol\right);n_{Cl_2}=n_{MnO_2}=0,9\left(mol\right)\)

\(\Rightarrow m_{HCl}=3,6.36,5=131,4\left(g\right)\)

Mặt khác, C% _{dung dịch HCl} = \(\dfrac{m_{HCl}}{m_{ddHCl}}.100\%\) \(\Leftrightarrow20\%=\dfrac{131,4}{m_{ddHCl}}.100\%\Leftrightarrow m_{ddHCl}=\dfrac{131,4.100}{20}=657\left(g\right)\)(\(m_{ddHCl}\) là khối lượng dung dịch HCl).

\(V_{Cl_2}=0,9.22,4=20,16\left(l\right)\)

b, Theo phương trình, \(n_{MnCl_2}=n_{MnO_2}=0,9\left(mol\right)\)

\(\Rightarrow m_{MnCl_2}=0,9.126=113,4\left(g\right)\)

Mặt khác,

m_{dung dịch sau phản ứng} = \(m_{MnCl_2}+\) m_{dung dịch HCl} - \(m_{Cl_2}\)

= \(78,3+657-\left(0,9.35,5.2\right)\)

= 671,4 (g)

\(\Rightarrow C\%\)_{dung dịch \(MnCl_2\)} = \(\dfrac{113,4}{671,4}.100\%=16,89\%\)

c, Theo (1), \(n_{Cl_2}=n_{MnCl_2}=0,9\left(mol\right)\)

PTHH: \(Fe+\dfrac{3}{2}Cl_2\rightarrow FeCl_3\left(2\right)\)

Số mol: 0,6 \(\rightarrow\) 0,9 \(\rightarrow\) 0,6

Theo (2) \(\Rightarrow n_{FeCl_3}=n_{Fe}=0,6\left(mol\right)\)

\(\Rightarrow m_{FeCl_3}=0.6.162,5=97,5\left(g\right)\)

\(\Rightarrow m\) _{dung dịch muối thu được} = 97,5 + 52,5 =150 (g).

\(\Rightarrow C\%\)_{dung dịch \(FeCl_3\)}= \(\dfrac{97,5}{150}.100\%=65\%\)

Cl₂+2NaBr ->2NaCl+Br₂

  0,04 -0,08-------0,08-----0,04 mol

Cl₂+2NaI ->2NaCl+I₂

0,03--0,06---0,06-----0,03 mol

n _NaBr =0,8.0,1=0,08 mol

n _NaI=0,6.0,1=0,06 mol

=>V _Cl2=(0,04+0,03).22,4=1,568l

=>m _Br2=0,04 .160=6,4g

=>m _I2=0,03.254=7,62g

b)

NaCl+AgNO₃->AgCl+NaNO₃

0,14-----------------0,14 mol

=>m _AgCl=0,14.143,5=20,09g

c)2KMnO₄+16HCl->2KCl+2MnCl₂+5Cl₂+8H₂O

    0,028--------------------------------------0,07 mol

=>m _KMnO4=0,028.158=4,424g