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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
\(a,CO_2+Ba(OH)_2\to BaCO_3\downarrow+H_2O\\ b,n_{CO_2}=\dfrac{2,24}{22,4}=0,1(mol)\\ \Rightarrow n_{Ba(OH)_2}=0,1(mol)\\ \Rightarrow C_{M_{Ba(OH)_2}}=\dfrac{0,1}{0,2}=0,5M\\ c,n_{BaCO_3}=0,1(mol)\\ \Rightarrow m_{BaCO_3}=0,1.197=19,7(g)\)
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)
a,
\(28,4g\left\{{}\begin{matrix}X_2CO_3\\MgCO_3\end{matrix}\right.\)\(+HCl-->6,72l\)
\(n_{CO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
\(X_2CO_3+2HCl-->2XCl+H_2O+CO_2\)
\(Y_2CO_3+2HCl-->2YCl+H_2O+CO_2\)
Theo định luật bảo toàn khối lượng :
\(m_{M_2CO_3}+m_{HCl}=m_{muối}+m_{H_2O}+m_{CO_2}\)
\(28,4+21,9=m_{2.muối}+5,4+13,2\)
\(m_{2.muối}=31,7g\)
b,
\(M_X< M_{\overline{M}}< M_Y\)
Ta giải theo khối lượng mol trung bình :
\(M_2CO_3\left(0,3\right)+2HCl\left(0,6\right)-->2\overline{M}Cl\left(0,3\right)+H_2O\left(0,3\right)+CO_2\left(0,3\right)\)
\(m_{HCl}=0,6.36,5=21,9\left(g\right)\)
\(m_{CO_2}=0,3.44=13,2\left(g\right)\)
\(m_{H_2O}=0,3.18=5,4\left(g\right)\)
Theo định luật bảo toàn khối lượng :
\(m_{\overline{M_2}CO_3}=m_{\overline{M}Cl}+m_{H_2O}+m_{CO_2}-m_{HCl}\)
\(=31,7+5,4+13,2-21,9=38,4\left(g\right)\)
\(M_{\overline{M_2}CO_3}=\dfrac{m}{n}=\dfrac{28,4}{0,3}=94,67\left(g\right)\)
\(=>2M_{\overline{M}}+60=94,67\)
\(2M_{\overline{M}}=34,67\)
\(M_{\overline{M}}=17,335\left(g\right)\)
\(7\left(Li\right)< 17,335< 23\left(Na\right)\)
\(=>X:Li\)
\(Y:Na\)
\(c,CO_2\left(0,1\right)+Ba\left(OH\right)_2\left(0,05\right)-->Ba\left(HCO_3\right)_2\)
\(CO_2\left(0,2\right)+Ba\left(OH\right)_2\left(0,2\right)-->BaCO_3\downarrow\left(0,2\right)+H_2O\)
\(n_{BaCO_3}=\dfrac{39,4}{197}=0,2\left(mol\right)\)
\(n_{Ba\left(OH\right)_2}=0,2+0,05=0,25\left(mol\right)\)
\(C_{MBa\left(OH\right)_2}=\dfrac{0,25}{1,25}=2M\)