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a) Gọi số mol Fe, Mg là a, b (mol)
=> 56a + 24b = 1,04 (1)
\(n_{H_2}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\)
PTHH: Fe + 2HCl --> FeCl2 + H2
a--->2a-------------->a
Mg + 2HCl --> MgCl2 + H2
b---->2b------------->b
=> a + b = 0,03 (2)
(1)(2) => a = 0,01 (mol); b = 0,02 (mol)
mFe = 0,01.56 = 0,56 (g)
mMg = 0,02.24 = 0,48 (g)
b) nHCl(lý thuyết) = 2a + 2b = 0,06 (mol)
=> \(n_{HCl\left(tt\right)}=\dfrac{0,06.110}{100}=0,066\left(mol\right)\)
=> \(V_{dd.HCl\left(tt\right)}=\dfrac{0,066}{0,1}=0,66\left(l\right)\)
Fe+2HCl->Fecl2+H2
x-----2x-----------------x
Mg+2Hcl->MgCl2+h2
y-------2y----------------y
a)
ta có :\(\left\{{}\begin{matrix}56x+24y=10,2\\x+y=0,35\end{matrix}\right.\)
=>x=0,05625 , y=0,29375
=>m Fe=0,05625.56=3,15g
=>m Mg=7,05g
=>VHCl=\(\dfrac{0,05625.2+0,29375.2}{0,1}\)=7l=7000ml
1.
2M + 2xHCl \(\rightarrow\)2MClx + xH2
nH2=\(\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
Theo PTHH ta có:
nM=\(\dfrac{2}{x}\)nH2=\(\dfrac{0,6}{x}\)
MM=\(\dfrac{5,4}{\dfrac{0,6}{x}}=9x\)
Với x=3 thì MM=27
Vậy M là Al
Mg + 2HCl = MgCl2 +H2
x x
2Al + 6HCl= 2AlCl3 + 3H2
y y
2Cu + O2 = 2CuO
z z = 8/80 = 0,1 mol
3NaOH + AlCl3 = Al(OH)3 + 3NaCL
y y
Al(OH)3 + NaOH = NaALO2 + 2H2O
y y
2NaOH + MgCl2 = Mg(OH)2 + 2NaCl
x x
Mg(OH)2 = MgO + H2O
x x = 4/40 = 0,1 mol
=>mCu= 0,1*64=6,4
mMg=0,1*24=2,4
mAl=10-6,4-2,4=1,2
a. \(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
b. \(n_{H2}=\frac{1,12}{22,4}=0,05\left(mol\right)\)
\(\rightarrow n_{Fe}=n_{H2}=0,05\left(mol\right)\)
\(\rightarrow\%m_{Fe}=\frac{0,05.56}{10}=28\%,\%_{Fe2O3}=100\%-28\%=72\%\)
c. \(n_{Fe2O3}=0,045\left(mol\right)\)
\(\rightarrow n_{FeCl_3}0,045.2=0,09\left(mol\right)\)
\(n_{FeCl_2}=n_{Fe}=0,05\left(mol\right)\)
Ta có \(FeCl_2+2NaOH\rightarrow Fe\left(OH\right)_2+2NaCl\)
______0,05________________0,05
\(FeCl_3+3NaOH\rightarrow Fe\left(OH\right)_3+3NaCl\)
0,09________________0,09_____________
\(4Fe\left(OH\right)_2+O_2\rightarrow2Fe_2O_3+4H_2O\)
0,05______________0,025____________
\(3Fe\left(OH\right)_3\rightarrow Fe_2O_3+3H_2O\)
0,09______ 0,045______________
\(\rightarrow n_{Fe2O3}=0,025+0,045=0,07\left(mol\right)\)
\(\rightarrow m_{Fe2O3}=0,07.160=11,2\left(g\right)\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
x______________ x _______x
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
y____________ y________y
\(n_{H2}=\frac{4,48}{22,4}=0,2\left(mol\right)\)
\(\rightarrow x+y=0,2\left(1\right)\)
\(MgCl_2+2NaOH\rightarrow Mg\left(OH\right)_2+2NaCl\)
x____________________ x________________
\(FeCl_2+2NaOH\rightarrow2NaCl+Fe\left(OH\right)_2\)
y _____________________________y
\(Fe\left(OH\right)_2\rightarrow FeO+H_2O\)
y_____________y________
\(4FeO+O_2\rightarrow Fe_2O_3\)
y ____________1/2y
\(Mg\left(OH\right)_2\rightarrow MgO+H_2O\)
x_________ x_____________
Ta có mMgO + mFe2O3 =12
\(\Leftrightarrow40x+80y=12\left(2\right)\)
(1)(2) \(\rightarrow a=b=0,1\)\(m_{Mg}=0,1.24=2,4\left(g\right),m_{Fe}=0,1.56=5,6\left(g\right)\)
Cu ko tác dụng vs HCl nên m chất rắn ko tan là mCu
\(\rightarrow m_{Cu}=6,4\left(g\right)\)