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a, \(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
PTHH: Mg + 2HCl → MgCl2 + H2
Mol: 0,2 0,4 0,2 0,2
\(V_{H_2}=0,2.22,4=4,48\left(l\right)\)
b,\(C\%_{ddHCl}=\dfrac{0,4.36,5.100\%}{200}=7,3\%\)
c, mdd sau pứ = 4,8+200-0,2.2 = 204,4 (g)
\(C\%_{ddMgCl_2}=\dfrac{0,2.95.100\%}{204,4}=9,3\%\)
PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
Ta có: \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
Theo PT: \(\left\{{}\begin{matrix}n_{H_2}=n_{Zn}=0,2\left(mol\right)\\n_{HCl}=2n_{Zn}=0,4\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)
\(C_{M_{HCl}}=\dfrac{0,4}{0,2}=2\left(M\right)\)
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) \(Theopt:n_{HCl}=3n_{Al}=0,6mol\)
\(\Rightarrow C_Mdd_{HCl}=\dfrac{0,6}{0,2}=3M\)
PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
Ta có: \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,4\left(mol\right)\\n_{FeCl_2}=n_{H_2}=0,2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{HCl}=\dfrac{0,4}{2}=0,2\left(l\right)\\V_{H_2}=0,2\cdot22,4=4,48\left(l\right)\\C_{M_{FeCl_2}}=\dfrac{0,2}{0,2}=1\left(M\right)\end{matrix}\right.\)
a) Zn + 2HCl \(\rightarrow\) ZnCl2 + H2
nZn = 6,5/65 = 0,1 mol
THeo pt: nH2 = nZn = 0,1 mol
=> VH2 = 0,1.22,4 = 2,24 lít
b) THeo pt: nHCl = 2nZn = 0,2 mol
=> mHCl = 0,2 . 36,5 = 7,3g
=> C%HCl = \(\dfrac{7,3}{200}.100\%=3,65\%\)
`Mg + 2HCl -> MgCl_2 + H_2`
`0,15` `0,3` `0,15` `(mol)`
`n_[Mg]=[3,6]/24=0,15(mol)`
`a)V_[H_2]=0,15.22,4=3,36(l)`
`b)m_[HCl]=0,3.36,5=10,95(g)`
`c)`
`H_2 + CuO` $\xrightarrow{t^o}$ `Cu + H_2 O`
`0,15` `0,15` `(mol)`
`=>m_[Cu]=0,15.64=9,6(g)`
\(n_{Mg}=\dfrac{3,6}{24}=0,15\left(mol\right)\)
PTHH: Mg + 2HCl ---> MgCl2 + H2
0,15->0,3------------------>0,15
CuO + H2 --to--> Cu + H2O
0,15------>0,15
=> \(V_{H_2}=0,15.22,4=3,36\left(l\right)\\ m_{HCl}=0,3.36,5=10,95\left(g\right)\\ m_{Cu}=0,15.64=9,6\left(g\right)\)
Bài 1.
a/ PTHH: Zn + 2HCl ===> ZnCl2 + H2
b/ nZn = 32,5 / 65 = 0,5 mol
=> nH2 = nZn = 0,5 mol
=> VH2(đktc) = 0,5 x 22,4 = 11,2 lít
Bài 2/
a/ PTHH: 2Al + 6HCl ==> 2AlCl3 + 3H2
b/ nAl = 4,05 / 27 = 0,15 mol
=> nH2 = 0,225 mol
=> VH2(đktc) = 0,225 x 22,4 = 5,04 lít
=> nAlCl3 = 0,15 mol
=> mAlCl3 = 0,15 x 133,5 = 20,025 gam
a, Zn + 2HCl ---> ZnCl2 + H2
b, nZn=\(\dfrac{13}{65}=0,2mol\)
Ta có: 1 mol Zn ---> 1 mol H2
nên 0,2 mol Zn ---> 0,2 mol H2
VH2=0,2.22,4=4,48 mol
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, \(n_{HCl}=2n_{Mg}=0,4\left(mol\right)\Rightarrow V_{HCl}=\dfrac{0,4}{2}=0,2\left(l\right)=200\left(ml\right)\)