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1.
RCO3 -> RO + CO2
Áp dụng ĐLBTKL ta có:
mRCO3=mRO+mCO2
=>mCO2=10-5,6=4,4((g)\(\Leftrightarrow\)0,1(mol)
VCO2=22,4.0,1=2,24(lít)
Theo PTHH ta có:
nRCO3=nCO2=0,1(mol)
MRCO3=\(\dfrac{10}{0,1}=100\)
=>MR=100-60=40
=>R là Ca
4.
R + H2SO4 -> RSO4 + H2
nH2=0,5(mol)
Theo PTHH ta có:
nR=nH2=0,5(mol)
MR=\(\dfrac{12}{0,5}=24\)
=>R là Mg
1.1. Al + NaOH + H2O ==> NaAlO2 + 3/2H2
nH2(1)=3,36/22,4=0.15(mol)
=> nAl(1)= nH2(1):3/2= 0.15:3/2= 0.1(mol)
2.Mg + 2HCl ==> MgCl2 + H2
3.2Al + 6HCl ==> 2AlCl3 + 3H2
4.Fe + 2HCl ==> FeCl2 + H2
=> \(n_{H_2\left(2,3,4\right)}=\) 10.08/22.4= 0.45(mol)
=> nH2(3)=0.1*3/2=0.15(mol)
MgCl2 + 2NaOH ==> Mg(OH)2 + 2NaCl
AlCl3 + 3NaOH ==> Al(OH)3 + 3NaCl
FeCl2 + 2NaOH ==> Fe(OH)2 + 2NaCl
\(n_{N_2O}=\dfrac{3.36}{22.4}=0.15\left(mol\right)\)
\(n_{Al}=a\left(mol\right),n_{Mg}=b\left(mol\right)\)
\(m_A=27a+24b=12.6\left(g\right)\left(1\right)\)
Bào toàn e :
\(3a+2b=0.15\cdot8\left(2\right)\)
\(\left(1\right),\left(2\right):a=0.2,b=0.3\)
\(\%Al=\dfrac{0.2\cdot27}{12.6}\cdot100\%=42.85\%\)
\(\%Mg=57.15\%\)
\(a,Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\\ n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ b,n_{Fe}=n_{H_2}=0,2\left(mol\right)\\ \%m_{Fe}=\dfrac{0,2.56}{12,8}.100\%=87,5\%\\ \%m_{Fe_2O_3}=100\%-87,5\%=12,5\%\\ c,n_{Fe_2O_3}=\dfrac{12,8-11,2}{160}=0,01\left(mol\right)\\ n_{H_2SO_4}=n_{Fe}+3n_{Fe_2O_3}=0,2+3.0,01=0,23\left(mol\right)\\ V_{ddH_2SO_4}=\dfrac{0,23}{0,46}=0,5\left(M\right)\)
PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
2a_____4a______2a____2a (mol)
\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
a_____2a______a_____a (mol)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
b______3b______b______\(\dfrac{3}{2}\)b (mol)
Ta lập HPT: \(\left\{{}\begin{matrix}2a+a+\dfrac{3}{2}b=\dfrac{13,44}{22,4}=0,6\\24\cdot2a+56a+27b=15,8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,1\cdot56}{15,8}\cdot100\%=35,44\%\\\%m_{Al}=\dfrac{0,2\cdot27}{15,8}=34,18\%\\\%m_{Mg}=30,38\%\end{matrix}\right.\)
Theo các PTHH: \(n_{HCl}=2n_{H_2}=1,2\left(mol\right)\)
\(\Rightarrow m_{ddHCl}=\dfrac{1,2\cdot36,5}{200}\cdot100\%=21,9\%\)
Ta có:
\(n_{CO_2}=\frac{1.12}{22.4}=0.05\left(mol\right)\) \(\Rightarrow m_{CO_2}=0.05\times44=2.2\left(g\right)\)
\(X_2CO_3+2HCl\rightarrow2XCl+H_2O+CO_2\)
\(YCO_3+2HCl\rightarrow YCl_2+H_2O+CO_2\)
Ta thấy
\(n_{HCl}=2n_{CO_2}=2\times0.05=0.1\left(mol\right)\)
\(\Rightarrow\) \(m_{HCl}=0.1\times36.5=3.65\left(g\right)\)
\(n_{H_2O}=n_{CO_2}=0.05\left(mol\right)\)
\(\Rightarrow\) \(m_{H_2O}=0.05\times18=0.9\left(g\right)\)
Áp dụng định luật bảo toàn khối lượng ta được:
\(m_{XCl+YCl_2}=\left(5.95+3.65\right)-\left(2.2+0.9\right)=9.6-3.1=6.5\left(g\right)\)
a) \(Fe+2HCl->FeCl_2+H_2\) (1)
\(2M+2xHCl->2MCl_x+xH_2\) (2)
=> \(n_{HCl}=2.n_{H_2}=2.\dfrac{1,008}{22,4}=0,09\left(mol\right)\)
=> mHCl = 0,09.36,5 = 3,285 (g)
Theo ĐLBTKL: \(m_A+m_{HCl}=m_{Muối}+m_{H_2}\)
=> \(m_A=4,575+0,045.2-3,285=1,38\left(g\right)\)
b) Đặt số mol Fe, M là a, b
=> 56a + M.b = 1,38 (***)
(1)(2) => a+ 0,5bx = 0,045 (*)
\(\left\{{}\begin{matrix}\dfrac{46.n_{NO_2}+64.n_{SO_2}}{n_{NO_2}+n_{SO_2}}=50,5\\n_{NO_2}+n_{SO_2}=\dfrac{1,8816}{22,4}=0,084\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}n_{NO_2}=0,063\\n_{SO_2}=0,021\end{matrix}\right.\)
Fe0 - 3e --> Fe+3
a---->3a
M0 -xe --> M+x
b-->bx
N+5 +1e--> N+4
___0,063<-0,063
S+6 + 2e --> S+4
___0,042<-0,021
Bảo oàn e: 3a + bx = 0,105 (**)
(*)(**) => \(\left\{{}\begin{matrix}a=0,015\\bx=0,06=>b=\dfrac{0,06}{x}\end{matrix}\right.\)
(***) => 0,015.56 + \(M.\dfrac{0,06}{x}\) = 1,38
=> M = 9x (g/mol)
Xét x = 1 => M = 9(L)
Xét x = 2 => M = 18(L)
Xét x = 3 => M = 27(Al)