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a, \(Mg+2HCl\rightarrow MgCl_2+H_2\)
b, \(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
Theo PT: \(n_{H_2}=n_{Mg}=0,2\left(mol\right)\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)
c, \(n_{HCl}=2n_{Mg}=0,4\left(mol\right)\Rightarrow V_{HCl}=\dfrac{0,4}{2}=0,2\left(l\right)=200\left(ml\right)\)
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\
pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,2 0,4 0,2
\(V_{H_2}=0,2.22,4=4,48\left(l\right)\\
C\%_{HCl}=\dfrac{0,4.36,5}{200}.100\%=7,3\%\)
Fe+2Hcl->FeCl2+H2
0,1---------------------0,1
2H2+O2-to>2H2O
0,1--------------0,1
n Fe=0,1 mol
=>VH2=0,1.22,4=2,24l
c) m H2O=0,1.18.95%=1,71g
\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
PTHH:
Fe + 2HCl ---> FeCl2 + H2
0,1 0,1
2H2 + O2 --to--> 2H2O
0,1 0,1
\(\rightarrow\left\{{}\begin{matrix}V_{H_2}=0,1.22,4=2,24\left(l\right)\\m_{H_2O}=0,1.18.\left(100\%-5\%\right)=1,71\left(g\right)\end{matrix}\right.\)
\(n_{CuO}=\dfrac{16}{80}=0,2mol\)
\(CuO+2HCl\rightarrow CuCl_2+H_2O\)
0,2 0,4 0,2 0,2
\(C_{M_{HCl}}=\dfrac{n_{HCl}}{V_{HCl}}=\dfrac{0,4}{0,2}=2M\)
\(m_{HCl}=0,4\cdot36,5=14,6g\)
nCuO = 16/80 = 0,2 (mol)
PTHH: CuO + 2HCl -> CuCl2 + H2
Mol: 0,2 ---> 0,4 ---> 0,2 ---> 0,2
CMCuCl2 = 0,2/0,2 = 1M
mHCl = 0,4 . 36,5 = 14,6 (g)
a, \(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(CuO+2HCl\rightarrow CuCl_2+H_2O\)
b, \(n_{Fe}=\dfrac{2,8}{56}=0,05\left(mol\right)\)
Theo PT: \(n_{H_2}=n_{Fe}=0,05\left(mol\right)\Rightarrow V_{H_2}=0,05.22,4=1,12\left(l\right)\)
c, \(n_{CuO}=\dfrac{4}{80}=0,05\left(mol\right)\)
Theo PT: \(n_{HCl}=2n_{Fe}+2n_{CuO}=0,2\left(mol\right)\)
\(\Rightarrow C_{M_{HCl}}=\dfrac{0,2}{0,2}=1\left(M\right)\)
PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
Ta có: \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
Theo PT: \(\left\{{}\begin{matrix}n_{H_2}=n_{Zn}=0,2\left(mol\right)\\n_{HCl}=2n_{Zn}=0,4\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)
\(C_{M_{HCl}}=\dfrac{0,4}{0,2}=2\left(M\right)\)
n Zn= 19,5/65=0,3 (mol).
PTPƯ: Zn(0.3) + HCl(0.6) ----> ZnCl2(0.3) + H2(0,3)
mHCl=0,6.36.5=21.9(g)
a) C%HCl= 21.9/300.100%=7,3%
b) VH2=0,3.22,4=6,72(lít)
c) mH2=0,3.2=0,6(g)
mZnCl2=0,3.136=40,8(g)
mddZnCl2 =(19,5+300)-0,6=318,9(g)
C%=mZnCl2/mddZnCl2.100= 40,8/318,9.100=12,793%
`a)PTHH:`
`Zn + 2HCl -> ZnCl_2 + H_2`
`0,1` `0,2` `0,1` `0,1` `(mol)`
`n_[HCl]=0,2.1=0,2(mol)`
`=>m_[Zn]=0,1.65=6,5(g)`
`b)m_[dd HCl]=1,1.200=220(g)`
`=>C%_[ZnCl_2]=[0,1.136]/[6,5+220-0,1.2].100~~6%`
\(a,n_{HCl}=0,2.1=0,2\left(mol\right)\)
PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
0,1<--0,2------>0,1------->0,1
\(\rightarrow m_{Zn}=0,1.65=6,5\left(g\right)\)
\(b,m_{ddHCl}=200.1,1=220\left(g\right)\)
\(\rightarrow m_{dd}=220+6,5-0,1.2=226,3\left(g\right)\\ \rightarrow C\%_{ZnCl_2}=\dfrac{0,1.136}{226,3}.100\%=6\%\)
a) \(Pt:2Al+6HCl\rightarrow2AlCl_3+3H_2\)
b) \(n_{Al}=\dfrac{5,4}{27}=0,2mol\)
\(Theopt:n_{H_2}=\dfrac{3}{2}n_{Al}=0,3mol\)
\(\Rightarrow V_{H_2}=0,3.22,4=6,72lít\)
c) \(Theopt:n_{HCl}=3n_{Al}=0,6mol\)
\(\Rightarrow C_Mdd_{HCl}=\dfrac{0,6}{0,2}=3M\)