Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1)
\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)
PTHH: 2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
____0,1----->0,15
=> mH2SO4 = 0,15.98 = 14,7(g)
=> \(C\%=\dfrac{14,7}{250}.100\%=5,88\%\)
2)
\(n_{Na_2CO_3}=\dfrac{21,2}{106}=0,2\left(mol\right)\)
PTHH: Na2CO3 + 2HCl --> 2NaCl + CO2 + H2O
_______0,2------------------------------>0,2
=> VCO2 = 0,2.22,4 = 4,48(l)
3)
\(n_A=\dfrac{18,4}{M_A}\left(mol\right)\)
PTHH: 2A + Cl2 --to--> 2ACl
____\(\dfrac{18,4}{M_A}\)---------->\(\dfrac{18,4}{M_A}\)
=> \(\dfrac{18,4}{M_A}\left(M_A+35,5\right)=46,8=>M_A=23\left(Na\right)\)
4)
nHCl = 0,2.3 = 0,6(mol)
PTHH: M + 2HCl --> MCl2 + H2
____0,3<-----0,6
=> \(M_M=\dfrac{7,2}{0,3}=24\left(Mg\right)\)
a) 2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
b) \(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\)
PTHH: 2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
0,4--->0,6-------------------->0,6
=> VH2 = 0,6.22,4 = 13,44 (l)
c) \(V_{dd.H_2SO_4}=\dfrac{0,6}{1}=0,6\left(l\right)\)
d) \(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)
PTHH: Fe2O3 + 3H2 --to--> 2Fe + 3H2O
Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,6}{3}\) => Fe2O3 hết, H2 dư
PTHH: Fe2O3 + 3H2 --to--> 2Fe + 3H2O
0,1----------------->0,2
=> mFe = 0,2.56 = 11,2 (g)
a)
$Fe + H_2SO_4 \to FeSO_4 + H_2$
$n_{H_2SO_4} = n_{H_2} = n_{Fe} = \dfrac{16,8}{56} = 0,3(mol)$
$V = 0,3.22,4 = 6,72(lít)$
$C_{M_{H_2SO_4}} = \dfrac{0,3}{0,25} = 1,2M$
b)
$n_{CuO} = \dfrac{16}{80} = 0,2(mol)$
$CuO + H_2 \xrightarrow{t^o} Cu + H_2O$
$n_{CuO} < n_{H_2}$ nên $H_2$ dư
$n_{Cu} = n_{CuO} = 0,2(mol)$
$m_{Cu} = 0,2.64 = 12,8(gam)$
\(n_{Fe}=\dfrac{m}{M}=\dfrac{16,8}{56}=0,3\left(mol\right)\\ PT:Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
0,3 0,3 0,3 (mol)
a) V= n. 22,4 = 0,3 . 22,4 = 6,72(l)
\(C\%=\dfrac{m_{H_2SO_4}}{m_{dd}}.100\%=\dfrac{0,3.98}{200}.100\%=11,76\%\)
b) PT: \(H_2+CuO\underrightarrow{t^o}Cu+H_2O\)
0,3 0,3
=> mCu=n.M=0,3.64=19,2(g)
Ta có: \(n_{Fe}=\dfrac{14}{56}=0,25\left(mol\right)\)
a. \(PTHH:Fe+H_2SO_4--->FeSO_4+H_2\uparrow\)
b. Theo PT: \(n_{H_2}=n_{Fe}=n_{H_2SO_4}=0,25\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,25.22,4=5,6\left(lít\right)\)
c. Ta có: \(m_{H_2SO_4}=0,25.98=24,5\left(g\right)\)
\(\Rightarrow C_{\%_{H_2SO_4}}=\dfrac{24,5}{486,5}.100\%=5,04\%\)
\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
Pt : \(Fe+2HCl\rightarrow FeCl_2+H_2|\)
1 2 1 1
0,1 0,2 0,1 0,1
a) \(n_{H2}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)
\(V_{H2\left(dktc\right)}=0,1.22,4=2,24\left(l\right)\)
b) \(n_{HCl}=\dfrac{0,1.2}{1}=0,2\left(mol\right)\)
100ml = 0,1l
\(C_{M_{ddHCl}}=\dfrac{0,2}{0,1}=2\left(M\right)\)
c) \(n_{FeCl2}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)
\(C_{M_{FeCl2}}=\dfrac{0,1}{2}=0,05\left(M\right)\)
Chúc bạn học tốt
Mình xin lỗi bạn nhé , bạn sửa lại giúp mình chỗ :
\(C_{M_{FeCl2}}=\dfrac{0,1}{0,1}=1\left(M\right)\)
a)
$Fe + 2HCl \to FeCl_2 + H_2$
n H2 = n Fe = 11,2/56 = 0,2(mol)
V H2 = 0,2.22,4 = 4,48(lít)
b)
n HCl = 2n Fe = 0,2.2 = 0,4(mol)
=> CM HCl = 0,4/0,4 = 1M
c)
$CuO + H_2 \xrightarrow{t^o} Cu + H_2O$
Ta thấy :
n CuO = 64/80 = 0,8 > n H2 = 0,2 nên CuO dư
Theo PTHH :
n CuO pư = n Cu = n H2 = 0,2(mol)
n Cu dư = 0,8 - 0,2 = 0,6(mol)
Vậy :
%m Cu = 0,2.64/(0,2.64 + 0,6.80) .100% = 21,05%
%m CuO = 100% -21,05% = 78,95%
a)
\(Mg + H_2SO_4 \to MgSO_4 + H_2\\ n_{H_2} = n_{Mg} = \dfrac{3,6}{24} = 0,15(mol)\\ b)\\ CuO + H_2 \xrightarrow{t^o} Cu + H_2O\\ n_{Cu} = n_{H_2} = 0,15(mol)\\ \Rightarrow m_{Cu} = 0,15.64 = 9,6(gam)\)
\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ a.Mg+2HCl\rightarrow MgCl_2+H_2\\ n_{H_2}=n_{MgCl_2}=n_{Mg}=0,2\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\\ b.n_{HCl}=0,2.2=0,4\left(mol\right)\\ m_{ddHCl}=\dfrac{0,4.36,5.100}{7,3}=200\left(g\right)\\ c.m_{ddsau}=4,8+200-0,2.2=204,4\left(g\right)\\ C\%_{ddMgCl_2}=\dfrac{0,2.95}{204,4}.100\approx9,295\%\\ d.V_{ddHCl}=\dfrac{200}{1,05}=\dfrac{4000}{21}\left(ml\right)=\dfrac{4}{21}\left(l\right)\\ C_{MddHCl}=\dfrac{0,4}{\dfrac{4}{21}}=2,1\left(M\right)\)
a) \(n_{Fe}=\dfrac{1,12}{56}=0,02\left(mol\right)\)
PTHH: Fe + H2SO4 ---> FeSO4 + H2
0,02->0,02------------------>0,02
b) \(V_{H_2}=0,02.22,4=0,448\left(l\right)\)
c) \(C\%_{H_2SO_4}=\dfrac{0,02.98}{100}.100\%=1,96\%\)
d) \(n_{CuO}=\dfrac{0,8}{80}=0,01\left(mol\right)\)
PTHH: CuO + H2 --to--> Cu + H2O
LTL: 0,01 < 0,02 => H2 dư
Theo pthh: nCu = nCuO = 0,01 (mol)
=> mCu = 0,01.64 = 0,64 (g)