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1,7 gam chất không tan là Cu
=> \(\%m_{Cu}=\dfrac{1,7}{10}.100=17\%\)
Fe+ 2HCl ---------> FeCl2 + H2 ;
2Al + 6HCl ---------> 2AlCl3 + 3H2
Gọi x,y lần lượt là số mol Fe, Al
\(\left\{{}\begin{matrix}56x+27y=10-1,7\\x+\dfrac{3}{2}y=\dfrac{5,6}{22,4}\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\)
\(\%m_{Fe}=\dfrac{0,1.56}{10}.100=56\%\)
\(\%m_{Al}=\dfrac{0,1.27}{10}.100=27\%\)
\(Zn+2HCl->ZnCl_2+H_2\\ Mg+2HCl->MgCl_2+H_2\\ n_{Zn}=a\\ n_{Mg}=b\\ 65a+24b=11,3g\\ n_{H_2}=a+b=\dfrac{6,72}{22,4}=0,3\\ a=0,1\\ m_{Zn}=65.0,1=6,5g\)
2Al +6HCl ----->2AlCl3 +3H2(1)
Al2O3 +6HCl----->2AlCl3 +3H2O(2)
Ta có
n\(_{H2}=\frac{13,44}{22,4}=0,6\left(mol\right)\)
Theo pthh
n\(_{Al}=\frac{2}{3}n_{H2}=0,4\left(mol\right)\)
m\(_{Al}=0,4.27=10,8\left(g\right)\)
m\(_{Al2O3}=21-10,8=10,2\left(g\right)\)
Ta có
n\(_{Al2O3}=\frac{10,2}{102}=0,1mol\)
n\(_{HCl\left(2\right)}=6n_{Al}=0,6\left(mol\right)\)
n\(_{HCl\left(1\right)}=2n_{H2}=1,2mol\)
\(\in n_{HCl}=0,6+1,2=1,8mol\)
m\(_{HCl}=1,8.36,5=65,7\left(g\right)\)
m\(_{HCl\left(36\%\right)}=\frac{65,7.36\%}{100\%}=20,4\left(g\right)\)
V\(_{HCl}=20,4,1,18=24ml\)
Chúc bạn hok tốt
Đặt \(\left\{{}\begin{matrix}n_{Al}=x\\n_{Zn}=y\\n_{Cu}=z\end{matrix}\right.\) ( mol )
\(m_{hh}=27x+65y+64z=22,8\left(g\right)\) (1)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
x 1,5x ( mol )
\(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
y y ( mol )
\(n_{H_2}=1,5x+y=\dfrac{11,2}{22,4}=0,5\left(mol\right)\) (2)
B là Cu
\(2Cu+O_2\rightarrow\left(t^o\right)2CuO\)
z z ( mol )
\(n_{CuO}=z=\dfrac{5,5}{80}=0,06875\left(mol\right)\) (3)
\(\left(1\right);\left(2\right);\left(3\right)\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,2\\z=0,06875\end{matrix}\right.\)
\(\left\{{}\begin{matrix}m_{Al}=0,2.27=5,4\left(g\right)\\m_{Zn}=0,2.65=13\left(g\right)\\m_{Cu}=22,8-5,4-13=4,4\left(g\right)\end{matrix}\right.\)
\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
PTHH: Fe + 2HCl --> FeCl2 + H2
_____0,15<--------------------0,15
=> mFe = 0,15.56 = 8,4(g)
=> \(\left\{{}\begin{matrix}\%Fe=\dfrac{8,4}{24,4}.100\%=34,426\%\\\%Fe_2O_3=100\%-34,426\%=65,574\%\end{matrix}\right.\)
\(a.CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\\ n_{CaCO_3}=n_{CO_2}=\dfrac{448:1000}{22,4}=0,02\left(mol\right)\\ n_{HCl}=0,02.2=0,04\left(mol\right)\\ C\%_{ddHCl}=\dfrac{0,04.36,5}{1,18.200}\approx0,619\%\\b.m_{CaCO_3}=0,02.100=2\left(g\right)\\ \%m_{CaCO_3}=\dfrac{2}{5}.100=40\%\\ \%m_{CaSO_4}=100\%-40\%=60\% \)
Mình tra KLR của dd HCl trên mạng là 1,18g/ml nên áp dụng vào bài nha ^^
a, \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
Ta có :
\(n_{H2}=\frac{13,44}{22,4}=0,6\left(mol\right)\)
\(\rightarrow n_{AL}=\frac{2}{3}n_{H2}=0,4\left(mol\right)\)
\(\rightarrow m_{Al}=0,4.27=10,8\left(g\right)\)
\(\rightarrow\%m_{Al}=\frac{10,8}{21}.100\%=51,43\%\)
\(\%m_{Al2O3}=100\%-51,43\%=48,57\%\)
b, \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
\(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)
\(n_{HCl}=3n_{Al}+6n_{Al2O3}=1,8\left(mol\right)\)
\(\rightarrow m_{HCl}=18.36,5=65,7\left(g\right)\)
\(\rightarrow m_{dd_{HCl}}=\frac{65,7}{36}.100\%=182,5\left(g\right)\)
\(\rightarrow V_{HCl}=\frac{182,5}{1,18}=154,66\left(l\right)\)