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a)
$Zn + H_2SO_4 \to ZnSO_4 + H_2$
Theo PTHH : $n_{Zn} = n_{H_2} = \dfrac{2,8}{22,4} = 0,125(mol)$
$m_{Zn} = 0,125.65 = 8,125(gam)$
$m_{Cu} = 8,3 - 8,125 = 0,175(gam)$
$\%m_{Zn} = \dfrac{8,125}{8,3}.100\% = 97,9\%$
$\%m_{Cu} = 100\% -97,9\% = 2,1\%$
$n_{H_2SO_4} = n_{H_2} = 0,125(mol) \Rightarrow m_{H_2SO_4} = 0,125.98 = 12,25(gam)$
a) PHHH: Zn + H2SO4 -> ZnSO4 + H2
nH2=1,5(mol)
=>nZn=nH2=1,5(mol)
b) mZn= 1,5.65= 97,5(g)
=> mCu=200-97,5=102,5(g)
Gọi x,y là số mol của AI và Fe
2Al + 3H2SO4 -> Al2(SO4)3 + 3H2
x --------------------... \(\frac{3x}{2}\)
Fe + H2SO4 -> FeSO4 + H2
y ----------------------> y
n H2 = 0,56 / 22,4 = 0,025 mol
Ta có hệ \(\begin{cases}27x+56y=0,83\\x+\frac{3x}{2}=0,025\end{cases}\)
\(\begin{cases}x=0,01mol\\y=0,01mol\end{cases}\)
=> m Al = 0,01 x 27 = 0,27 g
=> m Fe = 0,01 x 56 = 0,56 g
=> % Al = 0,27 / 0,83 x 100% = 32,53 %
=> % Fe = 0,56 / 0,83 x 100% = 67,47 %
a. PTHH:
\(Zn+H_2SO_4--->ZnSO_4+H_2\)
\(Cu+H_2SO_4--\times-->\)
b. Theo PT: \(n_{Zn}=n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
\(\Rightarrow m_{_{ }Zn}=0,1.65=6,5\left(g\right)\)
\(\Rightarrow\%_{m_{Zn}}=\dfrac{6,5}{10,5}.100\%=61,9\%\)
\(\%_{m_{Cu}}=100\%-61,9\%=31,8\%\)
\(Zn + H_2SO_4 \rightarrow ZnSO_4 + H_2\)
Cu không phản ứng H2SO4 loãng nhé
\(n_{H_2}= \dfrac{2,24}{22,4}= 0,1 mol\)
Theo PTHH:
\(n_{Zn}=n_{H_2}= 0,1 mol\)
\(\Rightarrow m_{Zn}= 0,1 . 65= 6,5 g\)
\(\Rightarrow\)%mZn=\(\dfrac{6,5}{10,5} . 100\)%~ 61,9%
\(\Rightarrow\)%mCu= 100% - 61,9%=38,1 %
\(a.Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ b.n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\ n_{Zn}=n_{H_2}=0,1\left(mol\right)\\ \Rightarrow\%m_{Zn}=\dfrac{0,1.65}{10,5}.100=61,9\%\\ \%m_{Cu}=100-61,9=38,1\%\)
\(n_k=n_{H_2}=0,125\left(mol\right)\)
a,b, \(PTHH:Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
.............0,125...0,125....................0,125...
\(\Rightarrow m_{Fe}=7\left(g\right)\)
Do Cu không phản ứng với H2SO4 .
\(\Rightarrow m_{Cu}=m_{hh}-m_{Fe}=10-7=3\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%Fe=70\%\\\%Cu=30\%\end{matrix}\right.\)
c, Có : \(m_{dd}=m_{Fe}+m_{ddH_2SO_4}-m_{H_2}=206,75\left(g\right)\)
\(\Rightarrow C\%=\dfrac{m_{H_2SO_4}}{m_{dd}}.100\%\approx5,925\%\)
a, \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
b, Ta có: \(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
Theo PT: \(n_{Al}=\dfrac{2}{3}n_{H_2}=\dfrac{4}{15}\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{\dfrac{4}{15}.27}{10}.100\%=72\%\\\%m_{Cu}=28\%\end{matrix}\right.\)
c, Theo PT: \(n_{H_2SO_4}=n_{H_2}=0,4\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}=0,4.98=39,2\left(g\right)\)
\(\Rightarrow C\%_{H_2SO_4}=\dfrac{39,2}{300}.100\%\approx13,067\%\)
a) Gọi `n_{Al} = a (mol); n_{Fe} = b (mol)`
PTHH:
`2Al + 3H_2SO_4 -> Al_2(SO_4)_3 + 3H_2`
`Fe + H_2SO_4 -> FeSO_4 + H_`
b) `n_{H_2} = (0,56)/(22,4) = 0,025 (mol)`
Theo PT: `n_{H_2} = n_{Fe} + 3/2 n_{Al}`
`=> b + 1,5a = 0,025`
Giải hpt \(\left\{{}\begin{matrix}27a+56b=0,83\\1,5a+b=0,025\end{matrix}\right.\Leftrightarrow a=b=0,01\)
=> \(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,01.27}{0,83}.100\%=32,53\%\\\%m_{Fe}=100\%-32,53\%=67,47\%\end{matrix}\right.\)
Câu 4 :
Vì bạc không tác dụng với axit sunfuric loãng :
\(n_{H2}=\dfrac{0,224}{22,4}=0,01\left(mol\right)\)
a) Pt : \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2|\)
1 1 1 1
0,01 0,01 0,01
b) \(n_{Zn}=\dfrac{0,0.1}{1}=0,01\left(mol\right)\)
\(m_{Zn}=0,01.65=0,65\left(g\right)\)
\(m_{Ag}=1,73-0,65=1,08\left(g\right)\)
0/0Zn = \(\dfrac{0,65.100}{1,73}=37,57\)0/0
0/0Ag = \(\dfrac{1,08.100}{1,73}=62,43\)0/0
c) \(n_{ZnSO4}=\dfrac{0,01.1}{1}=0,01\left(mol\right)\)
⇒ \(m_{ZnSO4}=0,01.161=1,61\left(g\right)\)
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