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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, PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
b, Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
Theo PT: \(\left\{{}\begin{matrix}n_{HCl}=3n_{Al}=0,6\left(mol\right)\\n_{H_2}=\dfrac{3}{2}n_{Al}=0,3\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow m_{HCl}=0,6.36,5=21,9\left(g\right)\)
\(V_{H_2}=0,3.22,4=6,72\left(l\right)\)
c, Cách 1:
Theo PT: \(n_{AlCl_3}=n_{Al}=0,2\left(mol\right)\)
\(\Rightarrow m_{AlCl_3}=0,2.133,5=26,7\left(g\right)\)
Cách 2:
Ta có: \(m_{H_2}=0,3.2=0,6\left(g\right)\)
Theo ĐLBT KL, có: mAl + mHCl = mAlCl3 + mH2
⇒ mAlCl3 = mAl + mHCl - mH2 = 5,4 + 21,9 - 0,6 = 26,7 (g)
Bạn tham khảo nhé!
a, \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
b, \(n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\)
Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}=0,45\left(mol\right)\Rightarrow V_{H_2}=0,45.22,4=10,08\left(l\right)\)
c, Ta có: m dd sau pư = 8,1 + 200 - 0,45.2 = 207,2 (g)
Theo PT: \(n_{AlCl_3}=n_{Al}=0,3\left(mol\right)\)
\(\Rightarrow C\%_{AlCl_3}=\dfrac{0,3.133,5}{207,2}.100\%\approx19,33\%\)
\(n_{H_2}=\dfrac{14,874}{24,79}=0,6\left(mol\right)\)
PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
a, \(n_{Al}=\dfrac{2}{3}n_{H_2}=0,4\left(mol\right)\Rightarrow m_{Al}=0,4.27=10,8\left(g\right)\)
b, \(n_{HCl}=2n_{H_2}=1,2\left(mol\right)\Rightarrow m_{HCl}=1,2.36,5=43,8\left(g\right)\)
\(n_{H_2}=\dfrac{14,874}{24,79}=0,6\left(mol\right)\)
PTHH :
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
0,4 1,2 0, 4 0,6
\(m_{Al}=0,4.27=10,8\left(g\right)\)
\(m_{HCl}=1,2.36,5=43,8\left(g\right)\)
pứ: Fe + 2HCl -> FeCl2 + H2
b. nFe = \(\dfrac{5,6}{56}\)= 0,1 mol
Từ pt suy ra được: nHCl = 2.nFe= 0,2 mol
=> mHCl = 0,2. 36,5 = 7,3 g
c. nH2 = nFe = 0,1 mol
=> VH2 = 0,1.22,4 = 2,24 (lít)
`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
=> \(V_{H_2}=0,15.22,4=3,36\left(l\right)\\ m_{HCl}=0,3.36,5=10,95\left(g\right)\\ m_{Cu}=0,15.64=9,6\left(g\right)\)
`a)`
`Fe + H_2 SO_4 -> FeSO_4 + H_2`
`0,4` `0,4` `0,4` `(mol)`
`n_[Fe]=[22,4]/56=0,4(mol)`
`b)m_[FeSO_4]=0,4.152=60,8(g)`
`c)V_[H_2]=0,4.22,4=8,96(l)`
a, \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
b, \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}=0,3\left(mol\right)\Rightarrow V_{H_2}=0,3.22,4=6,72\left(l\right)\)
c, \(n_{AlCl_3}=n_{Al}=0,2\left(mol\right)\Rightarrow m_{AlCl_3}=0,2.133,5=26,7\left(g\right)\)