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a) \(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\)
PTHH: 2Al + 6HCl --> 2AlCl3 + 3H2
0,4---------------------->0,6
=> V = 0,6.22,4 = 13,44 (l)
b)
\(n_{Fe_3O_4}=\dfrac{29}{232}=0,125\left(mol\right)\)
Gọi số mol Fe3O4 pư là a (mol)
PTHH: Fe3O4 + 4H2 --to--> 3Fe + 4H2O
Xét tỉ lệ: \(\dfrac{0,125}{1}< \dfrac{0,6}{4}\) => Hiệu suất tính theo Fe3O4
PTHH: Fe3O4 + 4H2 --to--> 3Fe + 4H2O
a----------------->3a
=> 232(0,125-a) + 56.3a = 22,6
=> a = 0,1
=> \(H\%=\dfrac{0,1}{0,125}.100\%=80\%\)
nAl = 10,8 : 27 = 0,4 (mol)
pthh : Al + 6HCl-t--> AlCl3 + H2
0,4--->2,4 (mol)
=> V= VO2 = 2,4 . 22,4 = 53,76 ( l)
nFe3O4 = 29 : 232 = 0,125 (mol)
pthh Fe3O4 + 4H2 -t--> 3Fe+ 4H2O
0,125----------------->0,375 (mol)
nFe (tt ) = 22,6 : 56 = 0,403 (mol )
%H = 0,375 / 0,403 . 100 % = 93 %
a, nFe = 16,8/56 = 0,3 (mol)
PTHH: Fe + 2HCl -> FeCl2 + H2
Mol: 0,3 ---> 0,6 ---> 0,3 ---> 0,3
VH2 = 0,3 . 22,4 = 6,72 (l)
b, nCuO = 20/80 = 0,25 (mol)
PTHH: CuO + H2 -> (t°) Cu + H2O
LTL: 0,25 < 0,3 => H2 dư
Gọi nCuO (p/ư) = a (mol)
=> nCu (sinh ra) = a (mol)
Ta có: 80(0,25 - a) + 64a = 16,4
=> a = 0,225 (mol)
H = 0,225/0,25 = 90%
a, \(n_{Mg}=\dfrac{7,2}{24}=0,3\left(mol\right)\)
PTHH: Mg + H2SO4 ---> MgSO4 + H2
0,3--->0,3--------------------->0,3
=> mH2SO4 = 0,3.98 = 29,4 (g)
b, VH2 = 0,3.22,4 = 6,72 (l)
c, \(n_{Fe_2O_3}=\dfrac{32}{160}=0,2\left(mol\right)\)
PTHH: Fe2O3 + 3H2 --to--> 2Fe + 3H2O
LTL: \(0,2>\dfrac{0,3}{3}\) => Fe2O3 dư
Theo pthh: \(n_{Fe_2O_3\left(pư\right)}=\dfrac{1}{3}n_{H_2}=\dfrac{1}{3}.0,3=0,1\left(mol\right)\)
=> mFe2O3 (dư) = (0,2 - 0,1).160 = 16 (g)
\(n_{Mg}=\dfrac{7,2}{24}=0,3\left(mol\right)\\
pthh:Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
0,3 0,3 0,3
\(m_{H_2SO_4}=0,3.98=29,4g\\
V_{H_2}=0,3.22,4=6,72l\)
\(n_{Fe_2O_3}=\dfrac{32}{160}=0,2\left(mol\right)\\
pthh:Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
\(LTL:\dfrac{0,2}{1}>\dfrac{0,3}{3}\)
=> Fe dư
\(n_{Fe\left(p\text{ư}\right)}=\dfrac{1}{3}n_{H_2}=0,1\left(mol\right)\\
m_{Fe\left(d\right)}=\left(0,2-0,1\right).56=5,6g\)
\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ pthh:Mg+2HCl\rightarrow MgCl_2+H_2\)
0,2 0,4 0,2
\(\Rightarrow m_{HCl}=0,4.36,5=14,6\left(g\right)\\ V_{H_2}=0,2.22,4=4,48\left(l\right)\\ pthh:FeO+H_2\underrightarrow{t^o}Fe+H_2O\)
0,2 0,2 0,2
\(m_{Fe}=0,2.56=11,2\left(g\right)\)
a) Fe + H2SO4 --> FeSO4 + H2
b) \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
PTHH: Fe + H2SO4 --> FeSO4 + H2
0,1------------------------>0,1
=> VH2 = 0,1.22,4 = 2,24 (l)
c) \(n_{CuO}=\dfrac{12}{80}=0,15\left(mol\right)\)
PTHH: CuO + H2 --to--> Cu + H2O
Xét tỉ lệ: \(\dfrac{0,15}{1}>\dfrac{0,1}{1}\) => CuO dư, H2 hết
PTHH: CuO + H2 --to--> Cu + H2O
0,1<--0,1------->0,1
=> mCuO(dư) = (0,15 - 0,1).80 = 4 (g)
mCu = 0,1.64 = 6,4 (g)
a, PT: \(Fe+H_2SO_4\rightarrow H_2SO_4+H_2\)
b, Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
Theo PT: \(n_{H_2}=n_{Fe}=0,1\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)
c, Ta có: \(n_{CuO}=\dfrac{12}{80}=0,15\left(mol\right)\)
PT: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
Xét tỉ lệ: \(\dfrac{0,15}{1}>\dfrac{0,1}{1}\), ta được CuO dư.
Theo PT: \(n_{Cu}=n_{CuO\left(pư\right)}=n_{H_2}=0,1\left(mol\right)\)
\(\Rightarrow n_{CuO\left(dư\right)}=0,05\left(mol\right)\Rightarrow m_{CuO\left(dư\right)}=0,05.80=4\left(g\right)\)
\(m_{Cu}=0,1.64=6,4\left(g\right)\)
Bạn tham khảo nhé!
Bài 1.
\(n_{Fe}=\dfrac{5,6}{56}=0,1mol\)
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
0,1 0,1 0,1 0,1
\(V_{H_2}=0,1\cdot22,4=2,24l\)
\(n_{CuO}=\dfrac{12}{80}=0,15mol\)
\(CuO+H_2\rightarrow Cu+H_2O\)
0,15 0,1
\(\Rightarrow CuO\) dư và dư \(\left(0,15-0,1\right)\cdot80=4g\)
Bài 2.
\(n_P=\dfrac{3,1}{31}=0,1mol\)
\(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
0,1 0,125
\(V_{O_2}=0,125\cdot22,4=2,8l\)
\(n_{Zn}=\dfrac{19,5}{65}=0,3mol\)
\(2Zn+O_2\underrightarrow{t^o}2ZnO\)
0,3 0,125 0
0,25 0,125 0,25
0,05 0 0,25
\(\Rightarrow ZnO\) dư và dư \(0,05\cdot81=4,05g\)
Bài 1.
a, \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
PTHH: Fe + H2SO4 ---> FeSO4 + H2
Mol: 0,1 0,1
b, \(V_{H_2}=0,1.22,4=2,24\left(l\right)\)
c, \(n_{CuO}=\dfrac{12}{80}=0,15\left(mol\right)\)
Ta có: \(\dfrac{0,15}{1}>\dfrac{0,1}{1}\) ⇒ CuO dư, H2 hết
PTHH: CuO + H2 ---to----> Cu + H2O
Mol: 0,1 0,1
\(m_{CuOdư}=\left(0,15-0,1\right).80=4\left(g\right)\)
\(n_K=\dfrac{3.9}{39}=0.1\left(mol\right)\)
\(K+H_2O\rightarrow KOH+\dfrac{1}{2}H_2\)
\(0.1..................0.1......0.05\)
\(m_{KOH}=0.1\cdot56=5.6\left(g\right)\)
\(V_{H_2}=0.05\cdot22.4=1.12\left(l\right)\)
\(n_{CuO}=\dfrac{20}{80}=0.25\left(mol\right)\)
\(CuO+H_2\underrightarrow{^{^{t^0}}}Cu+H_2O\)
\(1.........1\)
\(0.25.......0.05\)
\(LTL:\dfrac{0.25}{1}>\dfrac{0.05}{1}\Rightarrow CuOdư\)
\(m_Z=m_{Cu}+m_{CuO\left(dư\right)}=0.05\cdot64+\left(0.25-0.05\right)\cdot80=19.2\left(g\right)\)
a) PTHH : \(FeO+H_2-t^o->Fe+H_2O\)
\(CuO+H_2-t^o->Cu+H_2O\)
Đặt \(\hept{\begin{cases}n_{FeO}=x\left(mol\right)\\n_{CuO}=y\left(mol\right)\end{cases}}\) => \(72x+80y=11,2\left(I\right)\)
Có : \(m_{O\left(lấy.đi\right)}=m_{giảm}=1,92\left(g\right)\)
=> \(n_{O\left(lấy.đi\right)}=\frac{1,92}{16}=0,12\left(mol\right)\) Vì H% = 80% => Thực tế : \(n_{O\left(hh\right)}=\frac{0,12}{80}\cdot100=0,15\left(mol\right)\)
BT Oxi : \(x+y=0,15\left(II\right)\)
Từ (I) và (II) suy ra : \(\hept{\begin{cases}x=0,1\\y=0,05\end{cases}}\)
=> \(\hept{\begin{cases}m_{FeO}=7,2\left(g\right)\\m_{CuO}=4\left(g\right)\end{cases}}\)
b) PTHH : \(Fe+H_2SO_4-->FeSO_4+H_2\)
BT Fe : \(n_{Fe}=n_{FeO}=0,1\left(mol\right)\)
Theo pthh : \(n_{H_2}=n_{Fe}=0,1\left(mol\right)\)
=> \(V_{H_2}=2,24\left(l\right)\)
BT Cu : \(n_{Cu}=n_{CuO}=0,05\left(mol\right)\)
=> \(m_{CR\left(ko.tan\right)}=0,05\cdot64=3,2\left(g\right)\)
a.
\(n_{Al}=\dfrac{5,4}{27}=0,2mol\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
0,2 0,3 ( mol )
\(V_{H_2}=0,3.22,4=6,72l\)
\(n_{Fe_3O_4}=\dfrac{46,4}{232}=0,2mol\)
\(Fe_3O_4+4H_2\rightarrow\left(t^o\right)3Fe+4H_2O\)
\(\dfrac{0,2}{1}\) > \(\dfrac{0,3}{4}\) ( mol )
0,2 0,6 ( mol )
a là Sắt ( Fe )
\(m_{Fe}=0,6.56=33,6g\)
chuẩn