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Ta có:
\(m_{HCl}=\frac{20.124,1}{100}=24,82\left(g\right)\)
\(\Rightarrow n_{HCl}=\frac{24,82}{36,5}=0,68\left(mol\right)\)
Gọi a,b lần lượt là số mol Al,CaCO3 phản ứng
\(PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\)
_______a ______3a________a _________1,5a
\(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\)
b___________2b ______ b_______b
a)Giải hệ PT:
\(\left\{{}\begin{matrix}3a+2b=0,68\left(1\right)\\3a+44b=7,4\left(2\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=0,12\\b=0,16\end{matrix}\right.\)
\(\Rightarrow m_{Al}=0,12.27=3,24\left(g\right)\)
\(\%m_{Al}=\frac{3,24.100}{3,24+16}=16,84\%\)
\(\%_{CaCO3}=100\%-16,84\%=83,16\%\)
b)\(m_{Dd\left(spu\right)}=3,24+16+124,1-7,4=135,94\left(g\right)\)
\(C\%_{AlCl3}=\frac{0,12.133,5.100}{135,94}=11,78\%\)
\(C\%_{CaCl2}=\frac{0,16.111.100}{135,94}=13,06\%\)
mình cần giải gấp khoảng 1 tiếng ai có thể giải giùm mình được ko
a) \(n_{AlCl_3}=\dfrac{6,675}{133,5}=0,05\left(mol\right)\)
PTHH: 2Al + 6HCl --> 2AlCl3 + 3H2
0,05<-----------0,05---->0,075
=> \(\%Al=\dfrac{0,05.27}{14,15}.100\%=9,54\%\)
=> \(\%Cu=\dfrac{14,15-0,05.27}{14,15}.100\%=90,46\%\)
b) \(V_{H_2}=0,075.22,4=1,68\left(l\right)\)
c) \(n_{Cu}=\dfrac{14,15-0,05.27}{64}=0,2\left(mol\right)\)
PTHH: 4Al + 3O2 --to--> 2Al2O3
0,05->0,0375
2Cu + O2 --to--> 2CuO
0,2-->0,1
=> \(V_{O_2}=\left(0,1+0,0375\right).22,4=3,08\left(l\right)\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\\ m_{AlCl_3}=6,675\left(mol\right)\\ n_{AlCl_3}=\dfrac{6,675}{133,5}=0,05\left(mol\right)\\ \Rightarrow n_{Al}=n_{AlCl_3}=0,05\left(mol\right)\\ \Rightarrow m_A=0,05.27=1,35\left(g\right);m_{Cu}=14,15-1,35=12,8\left(g\right)\\ \%m_{Cu}=\dfrac{12,8}{14,15}.100\approx90,459\%\\ \Rightarrow\%m_{Al}\approx9,541\%\\ b,n_{Cu}=\dfrac{12,8}{64}=0,2\left(mol\right)\\ n_{H_2}=\dfrac{3}{2}.n_{Al}=\dfrac{3}{2}.0,05=0,075\left(mol\right)\\ \Rightarrow V=V_{H_2\left(đktc\right)}=0,075.22,4=1,68\left(l\right)\\ 4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\\ 2Cu+O_2\rightarrow\left(t^o\right)2CuO\\ n_{O_2}=\dfrac{3}{4}.n_{Al}+\dfrac{1}{2}.n_{Cu}=\dfrac{3}{4}.0,05+\dfrac{1}{2}.0,2=0,0875\left(mol\right)\)
\(\Rightarrow V_{O_2\left(đktc\right)}=0,0875.22,4=1,96\left(l\right)\)
NaBr xmol NaI ymol+Br2
2NaI+Br2=>2NaBr +I2
=>y=m/(127-80)=m/47 mol
NaBr (x+y )mol +Cl2=>NaCl+Cl2
=>x+y=m/(80-35.5)=m/44.5=>x=5m/4183
=>mNaBr=515m/4183 mNaI=150m/47 =>%NaBr=3.71%
NaBr xmol NaI ymol+Br2
2NaI+Br2=>2NaBr +I2
=>y=m/(127-80)=m/47 mol
NaBr (x+y )mol +Cl2=>NaCl+Cl2
=>x+y=m/(80-35.5)=m/44.5=>x=5m/4183
=>mNaBr=515m/4183 mNaI=150m/47 =>%NaBr=3.71%
\(n_{H_2}=\dfrac{7,84}{22,4}=0,35mol\)
Gọi \(\left\{{}\begin{matrix}n_{Fe}=x\\n_{Zn}=y\end{matrix}\right.\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
x x ( mol )
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
y y ( mol )
Ta có:
\(\left\{{}\begin{matrix}56x+65y=21,4\\x+y=0,35\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,15\\y=0,2\end{matrix}\right.\)
\(\Rightarrow m_{Fe}=0,15.56=8,4g\)
\(\Rightarrow m_{Zn}=0,2.65=13g\)
\(\%m_{Fe}=\dfrac{8,4}{21,4}.100=39,25\%\)
\(\%m_{Zn}=100\%-39,25\%=60,75\%\)
\(m_{FeCl_2}=0,15.127=19,05g\)
\(m_{ZnCl_2}=0,2.136=27,2g\)
\(Đặt:n_{MnO_2}=a\left(mol\right),n_{KMnO_4}=b\left(mol\right)\)
\(m_{hh}=87a+158b=37.96\left(g\right)\left(1\right)\)
\(n_{Cl_2}=\dfrac{10.08}{22.4}=0.45\left(mol\right)\)
\(2KMnO_4+16HCl\rightarrow2KCl+2MnCl_2+5Cl_2+8H_2O\)
\(MnO_2+4HCl\rightarrow MnCl_2+Cl_2+2H_2O\)
\(n_{Cl_2}=a+2.5b=0.45\left(mol\right)\left(2\right)\)
\(\left(1\right),\left(2\right):a=0.4,b=0.02\)
\(\%MnO_2=\dfrac{0.4\cdot87}{37.96}\cdot100\%=91.68\%\\\%KMnO_4=100-91.68=8.32\% \)
\(m_M=m_{KCl}+m_{MnCl_2}=0.02\cdot74.5+\left(0.4+0.02\right)\cdot126=54.41g\)
2. Hòa tan hoàn toàn 32,8 gam hỗn hợp gồm CuO và Fe tác dụng với dung dịch HCl thu được 65,1g muối. Tính % khối lượng mỗi chất trong hỗn hợp ban đầu ?
---------------------------------------Giải--------------------------------------
PTHH : \(CuO+2HCl\rightarrow CuCl_2+H_2O\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
Đặt x_nCuO; y_nFe . Ta có hệ : \(\left\{{}\begin{matrix}80x+56y=32,8\\135x+127y=65,1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,3\end{matrix}\right.\)
\(\Rightarrow\%m_{CuO}=\frac{0,2.80}{32,8}.100=48,78\%\)
\(\Rightarrow\%m_{Fe}=100-48,78=51,22\%\)
Câu 1: