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Sửa đề: 3,785 (l) → 3,7185 (l)
a, \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)
b, Ta có: \(n_{H_2}=\dfrac{3,7185}{24,79}=0,15\left(mol\right)\)
Theo PT: \(n_{Al}=\dfrac{2}{3}n_{H_2}=0,1\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,1.27}{40}.100\%=6,75\%\\\%m_{Al_2O_3}=93,25\%\end{matrix}\right.\)
c, \(n_{Al_2O_3}=\dfrac{40.93,25\%}{102}=\dfrac{373}{1020}\left(mol\right)\)
Theo PT: \(n_{HCl}=3n_{Al}+6n_{Al_2O_3}=\dfrac{212}{85}\left(mol\right)\)
\(\Rightarrow V_{HCl}=\dfrac{\dfrac{212}{85}}{2}=\dfrac{106}{85}\left(l\right)\approx1247,06\left(ml\right)\)
d, \(n_{AlCl_3}=n_{Al}+2n_{Al_2O_3}=\dfrac{212}{255}\left(mol\right)\)
\(\Rightarrow m_{AlCl_3}=\dfrac{212}{255}.133,5=\dfrac{9434}{85}\left(g\right)\)
e, \(C_{M_{AlCl_3}}=\dfrac{\dfrac{212}{255}}{\dfrac{106}{85}}=\dfrac{2}{3}\left(M\right)\)
a) 2Al + 2NaOH + 2H2O --> 2NaAlO2 + 3H2
b) \(\left\{{}\begin{matrix}\%Fe=\dfrac{8}{15}.100\%=53,33\%\\\%Al=\dfrac{15-8}{15}.100\%=46,67\%\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}n_{Fe}=x\left(mol\right)\\n_{Fe_2O_3}=y\left(mol\right)\end{matrix}\right.\Rightarrow56x+160y=4,8\left(1\right)\)
\(PTHH:Fe+CuSO_4\rightarrow FeSO_4+Cu\\ \Rightarrow n_{Cu}=n_{Fe}=a\left(mol\right)\\ Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\\ Cu+2FeCl_3\rightarrow CuCl_2+2FeCl_2\\ \Rightarrow n_{Cu}=\dfrac{1}{2}n_{FeCl_3}=n_{Fe_2O_3}=b\left(mol\right)\\ \Rightarrow n_{Cu\left(dư\right)}=a-b=\dfrac{3,2}{64}=0,05\left(mol\right)\left(2\right)\)
\(\left(1\right)\left(2\right)\Rightarrow\left\{{}\begin{matrix}a=\dfrac{8}{135}\\b=\dfrac{1}{108}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\%_{Fe}=\left(\dfrac{8}{135}\cdot56\right):4,8\cdot100\%\approx69,14\%\\\%_{Fe_2O_3}\approx30,86\%\end{matrix}\right.\)
\(b,n_{HCl}=6n_{Fe_2O_3}=\dfrac{1}{18}\approx0,06\left(mol\right)\\ \Rightarrow V_{dd_{HCl}}=\dfrac{0,06}{1}=0,06\left(l\right)\)
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 \(\left\{{}\begin{matrix}27a+56b=0,83\\1,5a+b=0,025\end{matrix}\right.\Leftrightarrow a=b=0,01\)
=> \(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,01.27}{0,83}.100\%=32,53\%\\\%m_{Fe}=100\%-32,53\%=67,47\%\end{matrix}\right.\)
a) PTHH: 2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
b) \(n_{H_2}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\)
PTHH: 2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
_____0,02<---0,03<---------------------0,03
=> \(\left\{{}\begin{matrix}\%Al=\dfrac{0,02.27}{2,16}.100\%=25\%\\\%Cu=100\%-25\%=75\%\end{matrix}\right.\)
c) mH2SO4 = 0,03.98 = 2,94 (g)
=> \(C\%\left(H_2SO_4\right)=\dfrac{2,94}{200}.100\%=1,47\%\)
a)\(n_{H_2}=\dfrac{3,36}{22,4}=0,15mol\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,15 0,3 0,15
\(m_{Zn}=0,15\cdot65=9,75\left(g\right)\)
\(\%m_{Zn}=\dfrac{9,75}{17,85}\cdot100\%=54,62\%\)
\(\%m_{ZnO}=100\%-54,62\%=45,38\%\)
b)\(m_{ZnO}=17,85-9,75=8,1\left(g\right)\Rightarrow n_{ZnO}=\dfrac{8,1}{81}=0,1mol\)
\(ZnO+2HCl\rightarrow ZnCl_2+H_2O\)
0,1 0,2
\(\Rightarrow\Sigma n_{HCl}=0,3+0,2=0,5mol\)
\(\Rightarrow V=\dfrac{0,5}{1}=0,5l=500ml\)
\(n_{H_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)
\(2A1+2NAOH+2H_2O-2NaA10_2+H_2O\)
\(AI_2O_3=2NaOH+2NaOHA10_2+H_2O\)
\(n_{AI}=\dfrac{2}{3}n_{H_2}=\dfrac{2}{3}.0,6=0,4\left(mol\right)\)
\(m_{AI}=27.0,4=10,8\left(gam\right);mAI_2O_3=31,2-10,8=20,4\left(gam\right)\)
Biết làm mỗi câu A
a. PTHH:
Cu + H2SO4 ---x--->
Mg + H2SO4 ---> MgSO4 + H2
b. Ta có: \(n_{H_2SO_4}=2.\dfrac{100}{1000}=0,2\left(mol\right)\)
Theo PT: \(n_{Mg}=n_{H_2SO_4}=0,2\left(mol\right)\)
\(\Rightarrow m_{Mg}=0,2.24=4,8\left(g\right)\)
\(\Rightarrow\%_{m_{Mg}}=\dfrac{4,8}{6}.100\%=80\%\)
\(\%_{m_{Cu}}=100\%-80\%=20\%\)
c. Theo PT: \(n_{H_2}=n_{Mg}=0,2\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,2.22,4=4,48\left(lít\right)\)
d. PTHH: \(Cu+2H_2SO_{4_đ}\overset{t^o}{--->}CuSO_4+SO_2+2H_2O\)
C1:
nH2= 0,15(mol)
PTHH: Fe + 2 HCl-> FeCl2 + H2
0,15_____0,3______0,15___0,15(mol)
-> mFe= 0,15.56=8,4(g)
-> mCuO= 10-8,4=1,6(g)
b) -> nCuO= 0,02(mol)
PTHH: CuO +2 HCl -> CuCl2 + H2O
0,02_________0,04(mol)
c) nHCl (tổng)= 0,34(mol)
=> CMddHCl= 0,34/0,2=1,7(M)
C2:
a) Mg +2 HCl -> MgCl2 + H2
nH2= 0,2(mol) -> nMg= nMgCl2= nH2=0,2(mol); nHCl=0,4(mol)
mMg= 0,2.24=4,8(g) -> mCu= 5,2(g)
=> %mMg=(4,8/10).100=48%
=>%mCu= 100%-48%=52%
b) mMgCl2= 0,2.95=19(g)
mHCl=0,4.36,5=14,6(g) -> mddHCl= ?? Không cho C% sao tính ta
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