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a)
$Fe + 2HCl \to FeCl_2 + H_2$
$Zn + 2HCl \to ZnCl_2 + H_2$
$2Al + 6HCl \to 2AlCl_3 + 3H_2$
$n_{HCl} = 0,45.2 = 0,9(mol)$
Theo PTHH :
$n_{H_2} = \dfrac{1}{2}n_{HCl} = 0,45(mol)$
$V_{H_2} = 0,45.22,4 = 10,08(lít)$
b)
Bảo toàn khối lượng :
$m_{muối} = 21,3 + 0,9.36,5 - 0,45.2 = 53,25(gam)$
a, \(n_{H_2}=n_C=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
PTHH: 2CH3COOH + Mg ---> (CH3COO)2Mg + H2
0,2<-------0,1<---------0,1<--------------0,1
\(\rightarrow\left\{{}\begin{matrix}m_{Mg}=0,1.24=2,4\left(g\right)\\m_{MgO}=8,4-2,4=6\left(g\right)\end{matrix}\right.\\ n_{MgO}=\dfrac{6}{40}=0,15\left(mol\right)\)
\(\rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{2,4}{8,4}.100\%=28,57\%\\\%m_{MgO}=100\%-28,57\%=71,53\%\end{matrix}\right.\)
b, PTHH: MgO + 2CH3COOH ---> (CH3COO)2Mg + H2O
0,15---->0,3----------------->0,15
=> \(\left\{{}\begin{matrix}m_{ddB}=8,4+\dfrac{\left(0,2+0,3\right).60}{9\%}-0,1.2=341,53\left(g\right)\\m_{\left(CH_3COO\right)_2Mg}=\left(0,15+0,1\right).142=35,5\left(g\right)\end{matrix}\right.\\ \Rightarrow C\%_{\left(CH_3COO\right)_2Mg}=\dfrac{35,5}{341,53}.100\%=10,4\%\)
Câu 1:
\(n_{H_2}=\dfrac{2.91362}{22.4}=0.13mol\)
\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
a a a a
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
2b 3b b 3b
Ta có: \(\left\{{}\begin{matrix}24a+54b=2.58\\a+3b=0.13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0.04\\b=0.03\end{matrix}\right.\)
\(m_{Mg}=0.04\times24=0.96g\)
\(m_{Al}=0.03\times2\times27=1.62g\)
\(V_{H_2SO_4}=\dfrac{0.04+3\times0.03}{0.5}=0.26l\)
Câu 2:
\(n_{H_2}=\dfrac{3.136}{22.4}=0.14mol\)
\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
a a a a
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
b b b b
Ta có: \(\left\{{}\begin{matrix}24a+56b=4.96\\a+b=0.14\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a=0.09\\b=0.05\end{matrix}\right.\)
\(m_{Mg}=0.09\times24=2.16g\)
\(m_{Fe}=0.05\times56=2.8g\)
\(C\%_{H_2SO_4}=\dfrac{0.14\times98\times100}{200}=6.86\%\)
Câu 3:
\(n_{H_2}=\dfrac{1.568}{22.4}=0.07mol\)
\(Ba+H_2SO_4\rightarrow BaSO_4+H_2\)
a a a a
\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
b b b b
Ta có: \(\left\{{}\begin{matrix}137a+24b=3.94\\a+b=0.07\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a=0.02\\b=0.05\end{matrix}\right.\)
\(m_{Ba}=0.02\times137=2.74g\)
\(m_{Mg}=0.05\times24=1.2g\)
\(CM_{H_2SO_4}=\dfrac{0.07}{0.1}=0.7M\)
2Fe + 3 Cl 2 → t ° 2Fe Cl 3 ( M FeCl 3 = 162,5 gam)
Cu + Cl 2 → t ° Cu Cl 2
Fe + 2HCl → Fe Cl 2 + H 2 ( M FeCl 2 = 127 gam)
n Fe = x mol
Theo đề bài và phương trình hóa học trên ta có:
127x = 25,4 => 0,2 mol
162,5x + 135y = 59,5. Thay x = 0,2 vào phương trình, ta có:
32,5 + 135y = 59,5 => y = 0,2
m FeCl 3 = 0,2 x 162,5 = 32,5g
m CuCl 2 = 0,2 x 135 = 27g
% m FeCl 3 = 32,5 : (32,5 + 27).100% = 54,62%
% m CuCl 2 = 100% - 54,62% = 45,38%
Mg+2HCl->MgCl2+H2
a..............................a(mol)
Fe+2HCl->FeCl2+H2
b............................b(mol)
=>nCu=3,2/64=0,05mol
=>%mCu=(3,2.100%)/11,2=28,6%
\(=>\left\{{}\begin{matrix}24a+56b=11,2-3,2\\a+b=0,2\end{matrix}\right.=>\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\)
=>mMg=24.0,1=2,4g=>%Mg=(2,4.100%)/11,2=21,4%
=>%Fe=100%-21,4%-28,6%=50%
b, MgCl2+2NaOH->Mg(OH)2+2NaCL
FeCl2+2NaOH->Fe(OH)2+2NaCl
=>m(kết tủa)=mMg(OH)2+mFe(OH)2
=0,1(58+90)=14,8g
a) mCu= m(k tan)= 3,2(g)
=> m(Mg, Fe)= 11,2- 3,2=8(g)
nH2= 4,48/22,4=0,2(mol)
PTHH: Mg + 2 HCl -> MgCl2 + H2
a______________2a__a______a(mol)
Fe + 2 HCl -> FeCl2 + H2
b____2b_____b_____b(mol)
Ta có hpt:
\(\left\{{}\begin{matrix}24a+56b=8\\a+b=0,2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}m_{Mg}=24.0,1=2,4\left(g\right)\\m_{Fe}=56.0,1=5,6\left(g\right)\end{matrix}\right.\)
=> %mMg= (2,4/11,2).100=21,429%
%mFe= (5,6/11,2).100=50%
=>%mCu= (3,2/11,2).100=28,571%
b/ MgCl2 + 2 NaOH -> Mg(OH)2 + 2 NaCl
0,1___________________0,1(mol)
FeCl2 + 2 NaOH -> Fe(OH)2 +2 NaCl
0,1__________________0,1(mol)
m(kt)=mMg(OH)2 + mFe(OH)2= 58.0,1+ 90.0,1= 14,8(g)
Ta có: 56nFe + 24nMg = 13,6 (1)
PT: \(Fe+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Fe+H_2\)
\(Mg+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Mg+H_2\)
Theo PT: \(n_{CH_3COOH}=2n_{Fe}+2n_{Mg}=\dfrac{100.36\%}{60}=0,6\left(mol\right)\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Fe}=0,2\left(mol\right)\\n_{Mg}=0,1\left(mol\right)\end{matrix}\right.\)
a, \(n_{H_2}=n_{Fe}+n_{Mg}=0,3\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,3.22,4=6,72\left(l\right)\)
b, \(n_{\left(CH_3COO\right)_2Fe}=n_{Fe}=0,2\left(mol\right)\Rightarrow m_{\left(CH_3COO\right)_2Fe}=0,2.174=34,8\left(g\right)\)
\(n_{\left(CH_3COO\right)_2Mg}=n_{Mg}=0,1\left(mol\right)\Rightarrow m_{\left(CH_3COO\right)_2Mg}=0,1.142=14,2\left(g\right)\)