Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
\(MgO+2HCl\rightarrow MgCl_2+H_2O\)
\(n_{Mg}=n_{H_2}=\dfrac{1.12}{22.4}=0.05\left(mol\right)\)
\(m_{Mg}=0.05\cdot24=1.2g\)
\(m_{MgO}=9.2-1.2=8\left(g\right)\)
\(n_{H2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
a) Pt : \(Zn+2HCl\rightarrow ZnCl_2+H_2|\)
1 2 1 1
0,2 0,2
b) \(n_{Zn}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)
\(m_{Zn}=0,2.65=13\left(g\right)\)
\(m_{Cu}=19,4-13=6,4\left(g\right)\)
Chúc bạn học tốt
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 \(\left\{{}\begin{matrix}27a+56b=0,83\\1,5a+b=0,025\end{matrix}\right.\Leftrightarrow a=b=0,01\)
=> \(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,01.27}{0,83}.100\%=32,53\%\\\%m_{Fe}=100\%-32,53\%=67,47\%\end{matrix}\right.\)
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, \(C_{M_{AlCl_3}}=\dfrac{\dfrac{212}{255}}{\dfrac{106}{85}}=\dfrac{2}{3}\left(M\right)\)
\(n_{H2}=\dfrac{4,2}{22,4}=0,1875\left(mol\right)\)
a) Pt : \(2Al+6HCl\rightarrow2AlCl_3+3H_2|\)
2 6 2 3
a 1,5a
\(Fe+2HCl\rightarrow FeCl_2+H_2|\)
1 2 1 1
b 1b
b) Gọi a là số mol của Al
b là số mol của Fe
\(m_{Al}+m_{Fe}=6,225\left(g\right)\)
⇒ \(n_{Al}.M_{Al}+n_{Fe}.M_{Fe}=6,225g\)
⇒ 27a + 56b = 6,225g
Theo phương trình : 1,5a + 1b = 01875(2)
Từ(1),(2) ta có hệ phương trình :
27a + 56b = 6,225g
1,5a + 1b = 0,1875
⇒ \(\left\{{}\begin{matrix}a=0,075\\b=0,075\end{matrix}\right.\)
\(m_{Al}=0,075.27=2,025\left(g\right)\)
\(m_{Fe}=0,075.56=4,2\left(g\right)\)
0/0Al = \(\dfrac{2,025.100}{6,225}=32,53\)0/0
0/0Fe = \(\dfrac{4,2.100}{6,225}=67,47\)0/0
Chúc bạn học tốt
a) Fe + 2HCl --> FeCl2 + H2
b) \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
PTHH: Fe + 2HCl --> FeCl2 + H2
_____0,15<--------------------0,15
=> mFe = 0,15.56 = 8,4 (g)
=> mCu = 15-8,4 = 6,6 (g)
c) Số nguyên tử Fe = 0,15.6.1023 = 0,9.1023
a.b.\(n_{CO_2}=\dfrac{5,6}{22,4}=0,25mol\)
\(2CH_3COOH+Na_2CO_3\rightarrow2CH_3COONa+CO_2+H_2O\)
0,5 0,25 ( mol )
\(m_{CH_3COOH}=0,5.60=30g\)
\(\%m_{CH_3COOH}=\dfrac{30}{45}.100=66,67\%\)
\(\%m_{C_2H_5OH}=100\%-66,67\%=33,33\%\)
c.\(m_{NaOH}=50.20\%=10g\)
\(n_{NaOH}=\dfrac{10}{40}=0,25mol\)
\(2NaOH+CO_2\rightarrow Na_2CO_3+H_2O\)
\(\dfrac{0,25}{2}\) < \(\dfrac{0,25}{1}\) ( mol )
0,25 0,125 ( mol )
\(m_{Na_2CO_3}=0,125.106=13,25g\)
a)
2CH3COOH + Na2CO3 --> 2CH3COONa + CO2 + H2O
b) \(n_{CO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
PTHH: 2CH3COOH + Na2CO3 --> 2CH3COONa + CO2 + H2O
0,5<-----------------------------------0,25
=> mCH3COOH = 0,5.60 = 30 (g)
=> mC2H5OH = 45 - 30 = 15 (g)
c) \(n_{NaOH}=\dfrac{50.20\%}{40}=0,25\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{n_{NaOH}}{n_{CO_2}}=\dfrac{0,25}{0,25}=1\) => Tạo muối NaHCO3
PTHH: NaOH + CO2 --> NaHCO3
0,25-------------->0,25
=> mNaHCO3 = 0,25.84 = 21 (g)
\(n_{H_2}=\dfrac{3,36}{22,4}=0,15mol\\ a)ZnO+H_2SO_4\rightarrow ZnSO_4+H_2O\\ Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
0,15 0,15 0,15 0,15
\(b)m_{Fe}=0,15.56=8,4g\\ m_{ZnO}=16,5-8,4=8,1g\\ c)n_{ZnO}=\dfrac{8,1}{81}=0,1mol\\ ZnO+H_2SO_4\rightarrow ZnSO_4+H_2O\)
0,1 0,1 0,1 0,1
\(V_{ddH_2SO_4}=\dfrac{0,15+0,1}{2}=0,125M\\ d)Fe+CuSO_4\rightarrow FeSO_4+Cu\)
0,15 0,15 0,15 0,15
\(m_{rắn}=m_{ZnO}+m_{Cu}=8,1+0,15.64=17,7g\)
2CH3COOH+Na2CO3->2CH3COONa+H2O+CO2
0,6-------------------------------------------------------------0,3
n CO2=0,3 mol
=>m CH3COOH=0,6.60=36g
=>m C2H5OH=4,4g