Ta có :

\(n_{Al}=\frac{5,4}{27}=0,2\left(mol\right)\)

\(PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\)

PƯ: _______0,2 ___0,6_____0,2_____________(mol)

\(a,m_{AlCl3}=0,2.\left(27+35,5.3\right)=26,7\left(g\right)\)

\(b,V_{HCl}=0,5.0,6=0,3\left(l\right)\)

Vì đề cho sau phản ứng chỉ còn 1 chất tan (KCl) vậy nên phản ứng này hết. Tính theo nHCl

\(KOH+HCl\rightarrow KCl+H_2O\)

0,1___0,1_________________

\(n_{HCl}=0,1.1=0,1\left(mol\right)\)

\(\Rightarrow CM_{KOH}=x=\frac{0,1}{0,2}=0,5M\)

\(Mg+2HCl-->MgCl2+H2\)

x-----------------------------------------x(mol)

\(2Al+6HCl-->2AlCl3+3H2\)

y-------------------------------------------1,5y(mol)

\(n_{H2}=\frac{7,728}{22,4}=0,345\left(mol\right)\)

Theo bài ta có hpt

\(\left\{{}\begin{matrix}24x+27y=6,93\\x+1,5y=0,345\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,12\\y=0,15\end{matrix}\right.\)

\(\%m_{Mg}=\frac{0,12.24}{6,93}.100\%=41,56\%\)

\(\%m_{Al}=100-41,56=58,44\%\)

\(NaX+AgNO3--->AgX+NaNO3\)

\(n_{NaX}=\frac{15,45}{23+X}\left(mol\right)\)

\(n_{AgX}=\frac{28,2}{108+X}\left(mol\right)\)

Theo pthh

\(n_{NaX}=n_{AgX}\Rightarrow\frac{15,45}{23+X}=\frac{28,2}{108+X}\)

\(\Rightarrow1668,6+15,45X=648,6+28,2X\)

\(\Rightarrow12,75X=1020\)

\(\Rightarrow X=80\left(Br\right)\)

Vậy CTHH: NaBr

a)\(2Al+6HCl-->2AlCl3+3H2\)

\(CuO+2HCl-->CuCl2+H2O\)

\(n_{H2}=\frac{3,36}{22,4}=0,15\left(mol\right)\)

\(n_{Al}=\frac{2}{3}n_{H2}=0,1\left(mol\right)\)

\(m_{Al}=0,1.27=2,7\left(g\right)\)

\(\%m_{Al}=\frac{2,7}{12,3}.100\%=21,95\%\)

\(\%m_{CuO}=100-21,95=78,05\%\)

b)\(n_{HCl\left(1\right)}=2n_{H2}=0,3\left(mol\right)\)

\(m_{CuO}=12,3-2,7=9,6\left(g\right)\)

\(n_{CuO}=\frac{9,6}{80}=0,12\left(mol\right)\)

\(n_{HCl\left(2\right)}=2n_{CuO}=0,24\left(mol\right)\)

\(\sum n_{HCl}=0,3+0,24=0,54\left(mol\right)\)

\(V_{HCl}=\frac{0,54}{1,5}=0,3\left(l\right)\)

2M + 2xHCl -> 2MCl_x + xH₂

Ta có:

n_M=n_MClx

\(\Leftrightarrow\frac{13}{M_M}=\frac{27,2}{M_M+35,5x}\)

với x=2=> M_M=65(t/m)

=>M là Zn

đáp án :C

M+2HCl=MCl+H2

=>m Cl2- =27,2-13=14,2

=> mol Cl2-=0,4

theo bảo toàn Cl- :2*mol HCl=1*mol Cl2-

=> 2*mol HCl=0,4 => mol M=0,2 => M=65(Zn)

Bài 8:

\(m_{HCl}=185,4.100\%=18,54\left(g\right)\)

Gọi x là số mol HCl thêm

\(\Rightarrow\frac{18,54+36,5x}{185,4+36,5x}=0,1657\)

\(\Rightarrow x=0,4\)

\(\Rightarrow V=0,4.22,4=8,96\left(l\right)\)

Bài 4:

a/ Ko tan là Cu

\(m_{AL,Fe}=11,5-3,2=8,3\left(g\right)\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

x_______3x________x _____1,5x

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

y ___2y_______ y_________y

\(n_{H2}=0,25\left(mol\right)\)

Đặt x, y là mol Al, Fe

\(\left\{{}\begin{matrix}27ax+56y=8,3\\1,5x+y=0,25\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\)

mAlCl3=...........

mFeCl2=............

b/

\(\Sigma n_{HCl}=3x+2y=0,3+0,2=0,5\left(mol\right)\)

\(m_{HCl}=0,5.26,5=18,25\left(g\right)\)

\(m_{dd_{HCl}}=\frac{18,25.100}{14,6}=125\left(g\right)\)

c/ %Al,Cu,Fe=............

d/

\(m_{dd}=11,25+125-3,2-\left(0,25.2\right)=132,55\left(g\right)\)

=>C% AlCl3,FeCl2=.....................

a)\(Mg+2HCl-->MgCl2+H2\)

x----------------------------------------x(mol)

\(2Al+6HCl-->2AlCl3+3H2\)

y----------------------------------------1,5y(mol)

\(n_{H2}=\frac{13,44}{22,4}=0,6\left(MOL\right)\)

theo bài ta có hpt

\(\left\{{}\begin{matrix}24x+27y=12,6\\x+1,5y=0,6\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,3\\y=0,2\end{matrix}\right.\)

\(\%m_{Mg}=\frac{0,3.24}{12,6}.100\%=57,14\%\)

\(\%m_{Al}=100-57,14=42,86\%\)

b)\(n_{HCl}=2n_{H2}=1,2\left(mol\right)\)

\(V_{HCl}=\frac{1,2}{2}=0,6\left(l\right)\)

c)\(m_{muối}=m_{MgCl2}+m_{AlCl3}=0,3.95+0,2.133,5=55,2\left(g\right)\)

\(PTHH:2Mg+2HCl\rightarrow MgCl_2+H_2\)

_________a_______ 2a _____a _______a

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

b ____3b______ b_________1,5b

\(n_{HCl}=0,5.1=0,5\left(mol\right)\)

\(n_{H2}=\frac{4,368}{22,4}=0,195\left(mol\right)\Rightarrow a+1,5b=0,195\)

\(m_{hh}=3,87\left(g\right)\Rightarrow24a+27b=3,87\)

\(\Rightarrow\left\{{}\begin{matrix}a=0,06\\b=0,09\end{matrix}\right.\)

\(\%m_{Mg}=\frac{24.0,06}{3,87}.100\%=37,21\%\)

\(\%m_{Al}=\frac{27.0,09}{3,87}.100\%=62,79\%\)

\(Al+3HCl\rightarrow AlCl_3+\frac{3}{2}H_2\)

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

BTKL: m dung dịch trước phản ứng + m kim loại = m dung dịch sau phản ứng + mH2

\(m_{tang}=m_{kl}-m_{H2}\)

\(\Rightarrow m_{H2}=11-10,2=0,8\left(g\right)\Rightarrow n_{H2}=\frac{0,8}{2}=0,4\left(mol\right)\)

Gọi số mol Al và Fe lần lượt là x, y

\(\left\{{}\begin{matrix}27x+56y=11\\1,5x+y=0,4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,4\end{matrix}\right.\)

Tự tính % nha

y =0,1 nha