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Sửa đề: Cho 5.4 g nhôm tác dụng hết với m (g) HCl. Hỗn hợp thu được sau phản ững hoà tan được tiếp với m'(g) Mg và thu được 2,24 l H2 ở đktc. Tìm m và m'
\(2Al\left(0,2\right)+6HCl\left(0,6\right)\rightarrow2AlCl_3+3H_2\)
\(Mg\left(0,1\right)+2HCl\left(0,2\right)\rightarrow MgCl_2+H_2\left(0,1\right)\)
\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
\(\Rightarrow n_{HCl}=0,6+0,2=0,8\left(mol\right)\)
\(\Rightarrow m=0,8.36,5=29,2\left(g\right)\)
\(\Rightarrow m'=0,1.24=2,4\left(g\right)\)
\(n_{Mg}=\dfrac{4,8}{24}=0,2mol\)
\(n_{H_2}=\dfrac{2,24}{22,4}=0,1mol\)
Mg+2HCl\(\rightarrow\)MgCl2+H2
\(n_{Mg\left(pu\right)}=n_{H_2}=0,1mol\rightarrow n_{Mg\left(dư\right)}=0,2-0,1=0,1mol\)
\(m_{Mg}=0,1.24=2,4gam\)
\(n_{MgCl_2}=n_{H_2}=0,1mol\)
\(m_{MgCl_2}=0,1.95=9,5gam\)
Câu 2 :
\(n_{Cu}=a\left(mol\right),n_{Al}=b\left(mol\right)\)
\(m=64a+27b=11.8\left(g\right)\left(1\right)\)
\(BTKL:m_{O_2}=18.2-11.8=6.4\left(g\right)\)
\(n_{O_2}=\dfrac{6.4}{32}=0.2\left(mol\right)\)
\(2Cu+O_2\underrightarrow{^{^{t^0}}}2CuO\)
\(4Al+3O_2\underrightarrow{^{^{t^0}}}2Al_2O_3\)
\(n_{O_2}=0.5a+0.75b=0.2\left(mol\right)\left(2\right)\)
\(\left(1\right),\left(2\right):a=0.1,b=0.2\)
\(\%Cu=\dfrac{0.1\cdot64}{11.8}\cdot100\%=54.23\%\)
Đặt nMg=a(mol); nAl=b(mol)
PTHH: Mg +2 HCl -> MgCl2 + H2
a________2a_______a_____a(mol)
2 Al + 6 HCl -> 2 AlCl3 +3 H2
b_____3b____b_____1,5b(mol)
Ta có hpt: \(\left\{{}\begin{matrix}24a+27b=7,8\\a+1,5b=0,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,2\end{matrix}\right.\)
=> %mMg=[(0,1.24)/7,8].100=30,769%
=>%mAl= 69,231%
c) MgCl2 + 2 NaOH -> Mg(OH)2 + 2 NaCl
0,1_______________0,1(mol)
AlCl3 + 3 NaOH -> Al(OH)3 + 3 NaCl
0,2____________0,2(mol)
=> m=m(kết tủa)= mMg(OH)2+ mAl(OH)3= 58.0,1+ 78.0,2= 21,4(g)
\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right);n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{H_2\left(tổng\right)}=\dfrac{3}{2}.n_{Al}+n_{Fe}=\dfrac{3}{2}.0,2+0,3=0,6\left(mol\right)\\ a,V=V_{H_2\left(đktc\right)}=0,6.22,4=13,44\left(l\right)\\ b,n_{HCl}=\dfrac{6}{2}.n_{Al}+2.n_{Fe}=\dfrac{6}{2}.0,2+2.0,3=1,2\left(mol\right)\\ \Rightarrow m_{HCl}=1,2.36,5=43,8\left(g\right)\\ c,n_{O_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\\ 2H_2+O_2\rightarrow\left(t^o\right)2H_2O\\ Vì:\dfrac{0,6}{2}>\dfrac{0,25}{1}\Rightarrow H_2dư,O_2hết\\ n_{H_2O}=2.n_{O_2}=2.0,25=0,5\left(mol\right)\\ \Rightarrow m_{H_2O}=0,5.18=9\left(g\right)\)
\(n_{Al}=\dfrac{5,4}{27}=0,2mol;n_{H_2}=\dfrac{2,24}{22,4}=0,1mol\)
2Al+6HCl\(\rightarrow\)2AlCl3+3H2(1)
3Mg+2AlCl3\(\rightarrow\)3MgCl2+2Al(2)
Mg+2HCl\(\rightarrow\)MgCl2+H2(3)
-Theo PTHH(1,2): \(n_{Mg}=\dfrac{3}{2}n_{AlCl_3}=\dfrac{3}{2}n_{Al}=\dfrac{3}{2}.0,2=0,3mol\)
-Theo PTHH(3): \(n_{Mg}=n_{H_2}=0,1mol\)
\(\rightarrow n_{Mg}=0,3+0,1=0,4mol\rightarrow m=m_{Mg}=0,4.24=9,6gam\)