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\(n_{H_2}=\dfrac{11,2}{22,4}=0,5mol\)
Gọi \(\left\{{}\begin{matrix}n_{Al}=x\\n_{Mg}=y\end{matrix}\right.\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
x 3/2x
\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
y y ( mol )
Ta có:
\(\left\{{}\begin{matrix}27x+24y=10,2\\\dfrac{3}{2}x+y=0,5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,2\end{matrix}\right.\)
\(\Rightarrow m_{Al}=0,2.27=5,4g\)
\(\Rightarrow m_{Mg}=0,2.24=4,8g\)
=> Chọn C
Cho hỗn hợp qua dung dịch \(H_2SO_4\) loãng chỉ có Fe tác dụng.
\(\Rightarrow n_{H_2}=\dfrac{2,24}{22,4}=0,1mol\)
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
0,1 0,1
\(m_{Fe}=0,1\cdot56=5,6g\)
\(\Rightarrow m_{Cu}=10-5,6=4,4g\)
Gọi x,y lần lượt là số mol của Al, Mg
nH2 = \(\dfrac{6,72}{22,4}\)=0,3mol
Pt: 2Al + 6HCl --> 2AlCl3 + 3H2
......x.........................................1,5x
.....Mg + 2HCl --> MgCl2 + H2
......y......................................y
Ta có hệ pt:
\(\left\{{}\begin{matrix}1,5x+y=0,3\\27x+24y=10,2\end{matrix}\right.\)=> số âm xem lại
\(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
PTHH: Fe + H2SO4 ---> FeSO4 + H2
0,25<--------------------------0,25
\(\Rightarrow m_{Fe}=0,25.56=14\left(g\right)\\ \Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{14}{32}.100\%=43,75\%\\\%m_{FeO}=100\%-43,75\%=56,25\%\end{matrix}\right.\)
Gọi x, y lần lượt là số mol Al, Fe
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ n_{H_2}=\dfrac{0,56}{22,4}=0,025\left(mol\right)\\ Tacó:\left\{{}\begin{matrix}27x+56y=0,83\\1,5x+y=0,025\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=0,01\\y=0,01\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}m_{Al}=0,27\left(g\right)\\m_{Fe}=0,56\left(g\right)\end{matrix}\right.\)
Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
Theo PT: \(n_{HCl}=2n_{H_2}=0,2\left(mol\right)\Rightarrow C_{M_{HCl}}=\dfrac{0,2}{0,1}=2\left(M\right)\)
\(n_{Fe}=n_{H_2}=0,1\left(mol\right)\Rightarrow m_{Fe}=0,1.56=5,6\left(g\right)\)
\(\Rightarrow m_{Cu}=20-5,6=14,4\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{5,6}{20}.100\%=28\%\\\%m_{Cu}=72\%\end{matrix}\right.\)
PTHH: \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)
Ta có: \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)=n_{Fe}\)
\(\Rightarrow n_{Fe_2O_3}=0,15\left(mol\right)\) \(\Rightarrow m_{Fe_2O_3}=x=0,15\cdot160=24\left(g\right)\)
\(n_K=\dfrac{11,7}{39}=0,3\left(mol\right)\)
PT: \(2K+2H_2O\rightarrow2KOH+H_2\)
Theo PT: \(n_{KOH}=n_K=0,3\left(mol\right)\Rightarrow m_{KOH}=0,3.56=16,8\left(g\right)\)
→ Đáp án: B
Bài 5:
mCu= 43,24% . 14,8\(\approx\) 6,4(g)
=>mFe\(\approx\) 14,8 - 6,4= 8,4(g)
=> nFe\(\approx\) 8,4/56\(\approx\) 0,15(mol)
PTHH: Fe + 2 HCl -> FeCl2 + H2
nH2=nFe \(\approx\) 0,15 (mol)
=> V(H2,đktc) \(\approx\) 0,15 . 22,4\(\approx\) 3,36(l)
Bài 6:
nH2= 4,368/22,4=0,195(mol)
Đặt: nMg=a(mol); nAl=b(mol) (a,b>0)
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:
\(\left\{{}\begin{matrix}24a+27b=3,87\\a+1,5b=0,195\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,06\\b=0,09\end{matrix}\right.\)
a) nH2SO4= 2a+3b=0,39(mol)
=> mH2SO4= 0,39.98=38,22(g)
b) m(muối)= mMgSO4 + mAl2(SO4)3= 120a+ 133,5b= 120.0,06+133,5.0,09= 19,215(g)
\(n_{H_2}=\dfrac{2,24}{22,4}=0,1mol\)
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
0,1 0,1 ( mol )
\(m_{Fe}=0,1.56=5,6g\)
\(\rightarrow m_{Cu}=10-5,6=4,4g\)
--> B