Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Giả sử: \(\left\{{}\begin{matrix}n_{Mg}=a\left(mol\right)\\n_{Zn}=b\left(mol\right)\\n_{Cu}=c\left(mol\right)\\n_{Al}=d\left(mol\right)\end{matrix}\right.\)
_ Khi tác dụng với HCl.
Ta có: \(n_{H_2}=\dfrac{10,08}{22,4}=0,45\left(mol\right)\)
Theo ĐLBT mol e, có: 2a + 2b + 3d = 0,45.2 ⇒ 2a + 2b + 3d = 0,9 (1)
_ Khi tác dụng với H2SO4 đặc nóng.
Ta có: \(n_{SO_2}=\dfrac{10,64}{22,4}=0,475\left(mol\right)\)
Theo ĐLBT mol e, có: 2a + 2b + 2c + 3d = 0,475.2
⇒ 2a + 2b - 2c + 3d = 0,95 (2)
Trừ 2 vế của (1) và (2), có: c = 0,025 (mol)
\(\Rightarrow\%m_{Cu}=\dfrac{0,025.64}{14,7}.100\%\approx10,88\%\)
Bạn tham khảo nhé!
Đặt \(\left\{{}\begin{matrix}n_{Mg}=a\left(mol\right)\\n_{Zn}=b\left(mol\right)\\n_{Cu}=c\left(mol\right)\\n_{Al}=d\left(mol\right)\end{matrix}\right.\)
Ta có: \(\left\{{}\begin{matrix}n_{H_2}=\dfrac{10,08}{22,4}=0,45\left(mol\right)\\n_{SO_2}=\dfrac{10,64}{22,4}=0,475\left(mol\right)\end{matrix}\right.\)
Bảo toàn electron: \(\left\{{}\begin{matrix}2a+2b+2c+3d=0,475\cdot2\\2a+2b+3d=0,45\cdot2\end{matrix}\right.\)
\(\Rightarrow2c=0,475\cdot2-0,45\cdot2=0,05\) \(\Rightarrow c=0,025\)
\(\Rightarrow\%m_{Cu}=\dfrac{0,025\cdot64}{14,7}\cdot100\%\approx10,88\%\)
1)
Fe + 2HCl --> FeCl2 + H2
Cu + 2H2SO4 --> CuSO4 + SO2 + 2H2O
2)
- Xét TN1:
\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
PTHH: Fe + 2HCl --> FeCl2 + H2
0,15<------------------0,15
=> mFe = 0,15.56 = 8,4 (g)
\(\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{8,4}{14,8}.100\%=56,757\%\\\%m_{Cu}=100\%-56,757\%=43,243\%\end{matrix}\right.\)
3)
- Xét TN2:
\(n_{Cu}=\dfrac{29,6.43,243\%}{64}=0,2\left(mol\right)\)
PTHH: Cu + 2H2SO4 --> CuSO4 + SO2 + 2H2O
0,2-------------------------->0,2
=> V = 0,2.22,4 = 4,48 (l)
\(n_{Fe}=a\left(mol\right),n_{Zn}=b\left(mol\right)\)
\(m=56a+65b=13.22\left(g\right)\left(1\right)\)
\(n_{H_2}=\dfrac{4.928}{22.4}=0.22\left(mol\right)\)
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
\(n_{H_2}=a+b=0.22\left(mol\right)\left(2\right)\)
\(\left(1\right),\left(2\right):\)
\(a=0.12\)
\(b=0.1\)
\(\text{Bảo toàn e : }\)
\(n_{Zn}+n_{Fe}=n_{SO_2}=\dfrac{0.12}{2}+\dfrac{0.1}{2}=0.11\left(mol\right)\)
\(V_{SO_2}=0.11\cdot22.4=2.464\left(l\right)\)
`2Fe + 6H_2 SO_[4(đ,n)] -> Fe_2(SO_4)_3 + 3SO_2 \uparrow + 6H_2 O`
`0,05` `0,15` `0,025` `(mol)`
`Cu + 2H_2 SO_[4(đ,n)] -> CuSO_4 + SO_2 \uparrow + 2H_2 O`
`0,225` `0,45` `0,225` `(mol)`
`n_[SO_2]=[6,72]/[22,4]=0,3(mol)`
Gọi `n_[Fe]=x` ; `n_[Cu]=y`
`=>` $\begin{cases} \dfrac{3}{2}x+y=0,3\\56x+64y=17,2 \end{cases}$
`<=>` $\begin{cases}x=0,05\\y=0,225 \end{cases}$
`@m_[Fe_2(SO_4)_3]=0,025.400=10(g)`
`@m_[CuSO_4]=0,225.160=36(g)`
`@m_[dd H_2 SO_4]=[(0,15+0,45).98]/80 .100=73,5(g)`
Sửa đề: 80% ---> 98% (80% chưa đặc nên không giải phóng SO2 được)
Gọi \(\left\{{}\begin{matrix}n_{Fe}=a\left(mol\right)\\n_{Cu}=b\left(mol\right)\end{matrix}\right.\)
\(\rightarrow56a+64b=17,2\left(1\right)\)
PTHH:
\(2Fe+6H_2SO_{4\left(đặc,nóng\right)}\rightarrow Fe_2\left(SO_4\right)_3+3SO_2\uparrow+6H_2O\)
a------>3a------------------->0,5a--------------->1,5a
\(Cu+2H_2SO_{4\left(đặc,nóng\right)}\rightarrow CuSO_4+SO_2\uparrow+2H_2O\)
b----->2b------------------->b------------->b
\(\rightarrow1,5a+b=\dfrac{6,72}{22,4}=0,3\left(2\right)\)
Từ \(\left(1\right)\left(2\right)\rightarrow\left\{{}\begin{matrix}a=0,05\left(mol\right)\\b=0,225\left(mol\right)\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}m_{Fe_2\left(SO_4\right)_3}=0,5.0,05.400=10\left(g\right)\\m_{CuSO_4}=0,225.160=36\left(g\right)\\m_{ddH_2SO_4}=\dfrac{\left(0,05.3+0,225.2\right).98}{98\%}=60\left(g\right)\end{matrix}\right.\)
a, \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ n_{SO_2}=\dfrac{10,08}{22,4}=0,45\left(mol\right)\)
PTHH:
Zn + H2SO4 (loãng) ---> ZnSO4 + H2
0,2<--------------------------------------0,2
Zn + 2H2SO4 (đặc) ---> ZnSO4 + SO2↑ + 2H2O
0,2--->0,4------------------------------->0,2
Cu + 2H2SO4 ---> CuSO4 + SO2↑ + 2H2O
0,2<--0,4<------------------------0,2
b, \(\left\{{}\begin{matrix}m_{Zn}=0,2.65=13\left(g\right)\\m_{Cu}=0,2.64=12,8\left(g\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{13}{13+12,8}.100\%=50,4\%\\\%m_{Cu}=100\%-50,4\%=49,6\%\end{matrix}\right.\)
c, PTHH:
SO3 + H2O ---> H2SO4
0,4<---------------0,4
2SO2 + O2 --to, V2O5--> 2SO3
0,4<---------------------------0,4
4FeS2 + 11O2 --to--> 2Fe2O3 + 8SO2
0,2<--------------------------------------0,4
=> \(m_{FeS_2}=\dfrac{0,2.120}{100\%-20\%}=30\left(g\right)\)
nH2=4,48/22,4=0,2 mol
Fe +2HCl -->FeCl2+H2
0,2 0,2 mol
=>mFe=0,2*56=11,2 g
nSO2=10,08/22,4=0,45 mol
gọi số mol của Cu là a mol
bảo toàn e ta có
Cu\(^0\)-->Cu\(^{+2}\)+2e
a 2a S\(^{+6}\) + 2e -->S\(^{+4}\)
Fe\(^0\)--> Fe\(^{+3}\)+3e 0,45 0,9
0,2 0,6
=>a=0,15=>mCu=0,15*64=9,6 g
=>mhh=9,6+11,2=20,8g
=>%Cu=9,6*100/20,8=46,15%
