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\(n_{Zn}=\dfrac{m}{M}=\dfrac{16,25}{65}=0,25\left(mol\right)\\ PTHH:Zn+H_2SO_4->ZnSO_4+H_2\)
tỉ lệ 1 : 1 : 1 : 1
n(mol) 0,25-->0,25------->0,25------>0,25
\(V_{H_2\left(dktc\right)}=n\cdot22,4=0,25\cdot22,4=5,6\left(l\right)\\ m_{ZnSO_4}=n\cdot M=0,25\cdot\left(65+32+16\cdot4\right)=40,25\left(g\right)\)
\(a) Zn + H_2SO_4 \to ZnSO_4 + H_2\\ b) n_{H_2} = n_{Zn} = \dfrac{13}{65}=0,2(mol)\\ V_{H_2} = 0,2.22,4 = 4,48(lít)\\ c) CuO + H_2 \xrightarrow{t^o} Cu + H_2O\\ n_{CuO} = n_{H_2} = 0,2(mol)\\ m_{CuO} = 0,2.80 = 16(gam)\)
a, \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
b, \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
Theo PT: \(n_{H_2}=n_{Zn}=0,2\left(mol\right)\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)
c, \(n_{Fe_3O_4}=\dfrac{18,56}{232}=0,08\left(mol\right)\)
PT: \(Fe_3O_4+4H_2\underrightarrow{t^o}3Fe+4H_2O\)
Xét tỉ lệ: \(\dfrac{0,08}{1}>\dfrac{0,2}{4}\), ta được Fe3O4 dư.
Theo PT: \(n_{Fe}=\dfrac{3}{4}n_{H_2}=0,15\left(mol\right)\Rightarrow m_{Fe}=0,15.56=8,4\left(g\right)\)
a) Zn + H2SO4 --> ZnSO4 + H2
b) \(n_{Zn}=\dfrac{1,95}{65}=0,03\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{1,47}{98}=0,015\left(mol\right)\)
PTHH: Zn + H2SO4 --> ZnSO4 + H2
Xét tỉ lệ: \(\dfrac{0,03}{1}>\dfrac{0,015}{1}\) => Zn dư, H2SO4 hết
PTHH: Zn + H2SO4 --> ZnSO4 + H2
____0,015<-0,015--->0,015->0,015
=> mZn(dư) = (0,03-0,015).65 = 0,975 (g)
c) VH2 = 0,015.22,4 = 0,336(l)
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right);n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\a, Zn+2HCl\rightarrow ZnCl_2+H_2\\ b,V\text{ì}:\dfrac{0,2}{1}>\dfrac{0,1}{1}\Rightarrow Zn\text{dư}\\ \Rightarrow n_{Zn\left(p.\text{ứ}\right)}=n_{ZnCl_2}=n_{H_2}=0,1\left(mol\right)\\b, m_{Zn\left(p.\text{ứ}\right)}=0,1.65=6,5\left(g\right)\\ n_{HCl}=0,1.2=0,2\left(mol\right)\\ m_{HCl}=0,2.36,5=7,3\left(g\right)\\ d,m_{ZnCl_2}=136.0,1=13,6\left(g\right)\)
\(a,PTHH:Zn+2HCl\to ZnCl_2+H_2\)
Phản ứng thế
\(b,n_{Zn}=\dfrac{1,3}{65}=0,02(mol)\\ \Rightarrow n_{ZnCl_2}=n_{H_2}=0,02(mol)\\ \Rightarrow m_{ZnCl_2}=0,02.136=2,72(g)\\ V_{H_2}=0,02.22,4=0,448(l)\)
a. \(n_{Zn}=\dfrac{6.5}{65}=0,1\left(mol\right)\)
PTHH : Zn + 2HCl -> ZnCl2 + H2
0,1 0,2 0,1
b. \(V_{H_2}=0,1.22,4=2,24\left(l\right)\)
c. \(m_{HCl}=0,2.36,5=7,3\left(g\right)\)
\(n_{Zn}=\dfrac{6,5}{65}=0,1mol\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,1 0,2 0,1
\(V_{H_2}=0,1\cdot22,4=2,24l\)
\(m_{HCl}=0,2\cdot36,5=7,3g\)
a, \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
b, \(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\)
Theo PT: \(n_{H_2}=n_{Zn}=0,3\left(mol\right)\Rightarrow V_{H_2}=0,3.22,4=6,72\left(l\right)\)
c, \(n_{ZnSO_4}=n_{Zn}=0,3\left(mol\right)\Rightarrow m_{ZnSO_4}=0,3.161=48,3\left(g\right)\)
d, \(n_{Fe_2O_3}=\dfrac{64}{160}=0,4\left(mol\right)\)
PT: \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
Xét tỉ lệ: \(\dfrac{0,4}{1}>\dfrac{0,3}{3}\), ta được Fe2O3 dư.
Mà: H% = 30% \(\Rightarrow n_{H_2\left(pư\right)}=0,3.30\%=0,09\left(mol\right)\)
Theo PT: \(\left\{{}\begin{matrix}n_{Fe}=\dfrac{2}{3}n_{H_2}=0,06\left(mol\right)\\n_{Fe_2O_3\left(pư\right)}=\dfrac{1}{3}n_{H_2}=0,03\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow n_{Fe_2O_3\left(dư\right)}=0,4-0,03=0,37\left(mol\right)\)
\(\Rightarrow a=m_{Fe}+m_{Fe_2O_3\left(dư\right)}=62,56\left(g\right)\)
a)
Zn + 2HCl → ZnCl2 + H2
b) nZn = \(\dfrac{3,5}{65}\)=\(\dfrac{7}{130}\) mol
Theo tỉ lệ phản ứng => nH2 = nZn= \(\dfrac{7}{130}\)mol
<=> V H2 = \(\dfrac{7}{130}\).22,4 = 1,206 lít
c) nZnCl2 = nZn => mZnCl2 = \(\dfrac{7}{130}\).136= 7,32 gam
Câu 1 Hoàn thành PTHH:
\(1)2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2\downarrow+O_2\uparrow\\ 2)FeO+2HCl\rightarrow FeCl_2+H_2O\\ 3)Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\\ 4)4P+5O_2\underrightarrow{t^o}2P_2O_5\)
Câu 2
\(n_{Zn}=\dfrac{m}{M}=\dfrac{13}{65}=0,2\left(mol\right)\)
a) \(PTHH:Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
1 1 1 1
0,2 0,2 0,2 0,2
b) \(V_{H_2}=n.24,79=0,2.24,79=4,958\left(l\right)\)
c) \(m_{H_2SO_4}=n.M=0,2.\left(2+32+16.4\right)=19,6\left(g\right).\)
câu 1
(1)2KMnO4➞(to)K2MnO4+MnO2+O2
(2)FeO+2HCl➞FeCl2+H2O
(3)Fe2O3+3H2SO4➞Fe2(SO4)3+3H2O
(4)4P+5O2➞2P2O5
Câu2
a)PTHH:Zn+H2SO4➞ZnSO4+H2
b)nZn=\(\dfrac{13}{65}\)=0,2(m)
n\(_{H_2SO_4}\)=\(\dfrac{24,5}{98}\)=0,25(m)
PTHH : Zn + H2SO4 ➞ ZnSO4 + H2
tỉ lệ :1 1 1 1
số mol
ban đầu :0,2 0,25
ta có tỉ lệ:\(\dfrac{0,2}{1}\)<\(\dfrac{0,25}{1}\)->H2SO4 dư
PTHH : Zn+ H2SO4 ➞ + ZnSO4+H2
tỉ lệ :1 1 1 1
số mol :0,2 0,2 0,2 0,2
v\(_{H_2}\)=0,2.22,4=4,48(l)
c)m\(_{ZnSO_4}\)=0,2.161=32,2(g)
m\(_{H_2}\)=0,2.2=0,4(g)