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Ta có: \(n_{CuSO_4}=0,3\left(mol\right)\)
a, PT: \(2Al+3CuSO_4\rightarrow Al_2\left(SO_4\right)_3+3Cu\)
______0,2____0,3_________________0,3 (mol)
b, \(m_{Al}=0,2.27=5,4\left(g\right)\)
c, \(m_{Cu}=0,3.64=19,2\left(g\right)\)
Bạn tham khảo nhé!
Câu 15 :
\(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\)
Pt : \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
0,4----->1,2------->0,4------>0,6
\(m_{HCl}=1,2.36,5=43,8\left(g\right)\)
\(\Rightarrow m_{ddHCl}=\dfrac{43.8.100\%}{25\%}=175,2\left(g\right)\)
\(m_{ddspu}=10,8+175,2-0,6.2=184,8\left(g\right)\)
\(C\%_{AlCl3}=\dfrac{0,4.133,5}{184,8}.100\%=28,9\%\)
\(2Al+3CuCl_2\rightarrow2AlCl_3+3Cu\\ n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ TheoPTHH:n_{AlCl_3}=n_{Al}=0,2\left(mol\right);n_{Cu}=\dfrac{3}{2}.n_{Al}=\dfrac{3}{2}.0,2=0,3\left(mol\right)\\ a,m_{Cu}=0,3.64=19,2\left(g\right)\\ b,m_{AlCl_3}=133,5.0,2=26,7\left(g\right)\)
Sửa đề: "Sợi dây nhôm có khối lượng là 16,2 g"
a) PTHH: \(2Al+3CuSO_4\rightarrow Al_2\left(SO_4\right)_3+3Cu\)
b) Ta có: \(n_{Al}=\dfrac{16,2}{27}=0,6\left(mol\right)\)
\(\Rightarrow n_{CuSO_4}=0,9mol\) \(\Rightarrow m_{ddCuSO_4}=\dfrac{0,9\cdot160}{25\%}=576\left(g\right)\)
c) Theo PTHH: \(n_{Cu}=n_{CuSO_4}=0,3mol\) \(\Rightarrow m_{Cu}=0,9\cdot64=57,6\left(g\right)\)
\(\left(a\right)2Al+3H_2O\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ \left(b\right)n_{H_2}=\dfrac{13,44}{22,4}=0,6mol\\ n_{Al}=\dfrac{0,6.2}{3}=0,4mol\\ m_{Al}=0,4.27=10,8g\\ \left(c\right)n_{O_2}=\dfrac{4,8}{32}=0,15mol\\ 4Al+3O_2\underrightarrow{t^0}2Al_2O_3\\ \Rightarrow\dfrac{0,4}{4}>\dfrac{0,15}{3}\Rightarrow Al.dư\\ n_{Al_2O_3}=\dfrac{0,15.2}{3}=0,1mol\\ m_{oxit}=m_{Al_2O_3}=0,1.102=10,2g\)
a: \(2Al+3H_2SO_4\rightarrow1Al_2\left(SO_4\right)_3+3H_2\uparrow\)
0,4 0,6 0,2 0,6
b: \(n_{H_2}=\dfrac{13.44}{22.4}=0.6\left(mol\right)\)
=>\(n_{Al}=0.4\left(mol\right)\)
\(m_{Al}=0.4\cdot27=10.8\left(g\right)\)
c: \(4Al+3O_2\rightarrow2Al_2O_3\)
0,4 0,2
\(m_{Al_2O_3}=0.2\left(27\cdot2+16\cdot3\right)=0.2\cdot102=20.4\left(g\right)\)
Đặt : \(n_{Al}=a\left(mol\right),n_{Fe}=b\left(mol\right)\)
\(m_{hh}=27a+56b=6.95\left(g\right)\left(1\right)\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
\(n_{H_2}=1.5a+b=\dfrac{3.92}{22.4}=0.175\left(mol\right)\left(2\right)\)
\(\left(1\right),\left(2\right):a=0.05,b=0.1\)
\(\%Al=\dfrac{0.05\cdot27}{6.95}\cdot100\%=19.42\%\)
\(\%Fe=100-19.42=80.58\%\)
\(n_{H_2}=\dfrac{3,92}{22,4}=0,175(mol)\\ a,PTHH:2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\\ Fe+H_2SO_4\to FeSO_4+H_2\\b,n_{Al}=x(mol);n_{Fe}=y(mol)\\ \Rightarrow 27x+56y=6,95;1,5x+y=0,175\\ \Rightarrow x=0,05(mol);y=0,1(mol)\\ \Rightarrow \%_{Al}=\dfrac{0,05.27}{6,95}.100\%=19,42\%\\ \Rightarrow \%_{Fe}=100\%-19,42\%=80,58\% \)
2Al+3H2SO4--------->Al2(SO4)3+3H2
Fe+H2SO4---------->FeSO4+H2
nH2=5,6/22,4=0,25 mol
Gọi nAl=x ,nFe=y
Cứ 2 mol al------> 3 mol H2
x-------->3x/2
Cứ 1 mol fe -----> 1 mol H2
y y
ta có hệ phương trình
27x+56y=8,3
3x/2+y=0,25
=>x=0,1 mol
y=0,1 mol
mAl=0,1.27=2,7 g
%mAl=2,7.100/8,3=32,5%
%mFe=100-32,5=67,5%
b. Theo Pthh thì tổng số mol của h2 bằng tổng số mol của H2SO4
nH2=nHCl=0,25
mHCl=0,25.36,5=9,125 g
mdung dịch =9,125.100/25=36,5 g
\(a.2Al+3CuCl_2\rightarrow2AlCl_3+3Cu\\b.n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right) \\ n_{Cu}=\dfrac{3}{2}n_{Al}=0,6\left(mol\right)\\\Rightarrow m_{Cu}=0,6.64=38,4\left(g\right)\\ n_{AlCl_3}=n_{Al}=0,4\left(mol\right)\\ \Rightarrow m_{AlCl_3}=0,4.133,5=53,4\left(g\right)\\ c.n_{CuCl_2}=\dfrac{3}{2}n_{Al}=0,6\left(mol\right)\\V_{CuCl_2}=\dfrac{0,6}{3}=0,2\left(l\right)\\ d.2Al+3Cl_2-^{t^o}\rightarrow2AlCl_3\\ n_{Cl_2}=\dfrac{3}{2}n_{AlCl_3}=0,6\left(mol\right)\\ \Rightarrow V_{Cl_2}=0,6.22,4=13,44\left(l\right)\\ n_{Al}=n_{AlCl_3}=0,4\left(mol\right)\\ \Rightarrow m_{Al}=0,4.27=10,8\left(g\right)\)