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nZn = 13/65 = 0,2 mol
Zn + H2SO4 -> ZnSO4 + H2
0,2 -> 0,2 -> 0,2 -> 0,2 mol
a/ VH2 = 0,2 * 22,4 = 4,48 l
b/ ZnSO4 = 0,2 * (65 + 32 + 16 *4) = 32,2 g
c/ CuO + H2 -> Cu + H2O
0,2 <- 0,2 -> 0,2 -> 0,2 (mol)
mCuO = 0,2 * (64 + 16) = 16 g
Ta có PTHH: Zn + H2SO4➝ ZnSO4 + H2
a) nZn = 13/65= 0,2 mol ⇒nH2= 0,2 mol⇒VH2=0,2.22.4=4.48l
b)nZnSO4= 0,2 mol⇒ mZnSO4= 0,2. (65+32+16.4)=32,2 g
c) Ta có PTHH: H2+CuO→Cu + H2O
nCuO= 0.2 mol⇒mCuO= 0,2.(16+64)=16g
1.
\(n_{Al}=\frac{5,4}{27}=0,2\left(mol\right);n_{H_2SO_4}=\frac{3.10^{23}}{6.10^{23}}=0,5\left(mol\right)\)
\(PTHH:2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
\(TL:\frac{0,2}{2}< \frac{0,5}{3}\) → H2SO4 dư
\(V_{H_2}=0,3.22,4=6,72\left(l\right)\\ m_{Al_2\left(SO_4\right)_3}=0,1.342=34,2\left(g\right)\)
2.
\(n_{H_2}=\frac{6,72}{22,4}=0,3\left(mol\right);n_{CuO}=\frac{32}{80}=0,4\left(mol\right)\)
\(PTHH:CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
\(TL:\frac{0,3}{1}< \frac{0,4}{1}\) → CuO dư
\(m_A=m_{Cu}=0,3.64=19,2\left(g\right)\)
1)
\(2Cu+O_2\rightarrow2CuO\)
a)\(n_{Cu}=\frac{2.56}{64}=0.04\left(mol\right)\)
\(\Rightarrow n_{CuO}=\frac{2}{2}\cdot n_{Cu}=\frac{2}{2}\cdot0.04=0.04\left(mol\right)\)
\(\Rightarrow m_{CuO}=0.04\cdot80=3.2\left(g\right)\)
2)
\(n_{CuO}=\frac{24}{80}=0.3\left(mol\right)\)
\(n_{Cu}=\frac{2}{2}\cdot n_{Cu}=\frac{2}{2}\cdot0.3=0.3\left(mol\right)\)
\(\Rightarrow m_{Cu}=0.3\cdot64=19.2\left(mol\right)\)
\(n_{O_2}=\frac{1}{2}\cdot n_{CuO}=\frac{1}{2}\cdot0.3=0.15\left(mol\right)\)
\(\Rightarrow m_O=0.15\cdot32=4.8\left(g\right)\)
a) \(3H_2SO_4+2Al\rightarrow Al_2\left(SO_4\right)_3+3H_2\) ( 1 )
\(H_2SO_4+Zn\rightarrow ZnSO_4+H_2\)( 2 )
b) \(n_{H_2\left(1\right)}=\dfrac{V}{22,4}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
\(\Rightarrow n_{Al}=n_{H_2\left(1\right)}\cdot\dfrac{2}{3}=0,3\cdot\dfrac{2}{3}=0,2\left(mol\right)\)
\(\Rightarrow m_{Al}=n\cdot M=0,2\cdot27=5,4\left(mol\right)\)
\(n_{H_2\left(2\right)}=\dfrac{V}{22,4}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
\(\Rightarrow n_{Zn}=n_{H_2\left(2\right)}=0,2\left(mol\right)\)
\(\Rightarrow m_{Zn}=n\cdot M=0,2\cdot65=13\left(g\right)\)
c) \(n_{H_2SO_4\left(1\right)}=n_{H_2\left(1\right)}=0,3\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4\left(1\right)}=n\cdot M=0,3\cdot98=29,4\left(g\right)\)
\(n_{H_2SO_4\left(2\right)}=n_{H_2\left(2\right)}=0,2\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4\left(2\right)}=n\cdot M=0,2\cdot98=19,6\left(g\right)\)
\(\Rightarrow m_{H_2SO_4}=m_{H_2SO_4\left(1\right)}+m_{H_2SO_4\left(2\right)}=29,4+19,6=49\left(g\right)\)
d) \(n_{Al_2\left(SO_4\right)_3}=n_{H_2\left(1\right)}\cdot\dfrac{1}{3}=0,3\cdot\dfrac{1}{3}=0,1\left(mol\right)\)
\(\Rightarrow m_{Al_2\left(SO_4\right)_3}=n\cdot M=0,1\cdot342=34,2\left(g\right)\)
\(n_{ZnSO_4}=n_{H_2\left(2\right)}=0,2\left(mol\right)\)
\(\Rightarrow m_{ZnSO_4}=n\cdot M=0,2\cdot161=32,2\left(g\right)\)
Bài 2:
2X + nH2SO4 -> X2(SO4)n + nH2
=> nX2(SO4)n = \(\frac{1}{2}n_X\)
=> \(\frac{34,2}{2X+96n}=\frac{1}{2}\cdot\frac{5,4}{X}\)
=> 5,4X + 259,2 n = 34,2X
=> 28,8X = 259,2n
=> X = 9n
=> n = 3
X = 27
X là Al
b) 2Al + 3H2SO4 -> Al2(SO4)3 + 3H2
=> nAl = 0,2 (mol)
=> nH2 = nH2SO4 = \(\frac{3}{2}n_{Al}\)= 0,3 (mol)
VH2 = 0,3 . 22,4 = 6,72 (lít)
Bài 1 :
nH2SO4 = 3*1023/6*1023 = 0.5 mol
Gọi: kim loại : A ( hóa trị n )
2A + nH2SO4 --> A2(SO4)n + nH2
1/n___0.5_________0.5/n______0.5
M = 12/1/n = 12n
BL :
n = 2 => M = 24 (Mg)
VH2 = 0.5*22.4 = 11.2 (l)
mMgSO4 = 0.5*120=60 g
Bài 2 :
Gọi: kim loại là B ( hóa trị n )
2B + 2nH2SO4 --> B2(SO4)n + nH2
2B________________2B+96n
5.4_________________34.2
<=> 34.2*2B = 5.4 ( 2B + 96n)
<=> 68.4B = 10.8B + 518.4n
<=> 57.6B = 518.4n
<=> B = 9n
BL :
n= 3 => B = 27 (Al)
VH2 = 0.3*22.4 = 6.72 (l)
nH2SO4 = 0.3 mol
Số phân tử H2SO4 là :
0.3*6*1023 = 1.8*1023 (phân tử)
Bài 1)
a \(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)
\(n_{Fe_2O_3}=\frac{4,8}{216}\approx\text{0,02 (mol)}\)
\(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)
0,02 0,06
\(m_{H_2SO_4}=98\cdot0,06=5,88\left(g\right)\)
b) \(m_{Fe_2\left(SO_4\right)_3}=0,02\cdot400=\text{290.24}\left(g\right)\)
Câu 2 mai làm
Câu 2
a)\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+H_2\)
\(n_{Al}=\frac{5,4}{2,7}=0,2\left(mol\right)\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+H_2\)
0,4 mol 0,6 mol 0,2 mol
\(V_{H_2}=0,2\cdot22,4=4,48\left(l\right)\)
b) \(m_{H_2SO_4}=0,6\cdot98=58,8\left(g\right)\)
Theo đề bài ta có : nH2 = \(\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
a) PTHH :
\(2Al+3H2SO4->Al2\left(SO4\right)3+3H2\uparrow\)
0,1mol......0,15mol.........0,05mol............0,15mol
b) Số nguyên tử nhôm tham gia pư là :
N = 0,1.6.10^23 = 0,6.10^23 ( ng tử )
Khối lượng nhôm tham gia pư là :
mAl = 0,1.27 = 2,7 (g)
c) khối lượng muốn nhôm sunfat tạo thành là :
mAl2(SO4)3 = 0,15. 342 = 51,3(g)
Câu 1
\(Fe+H2SO4-->FeSO4+H2\)
\(n_{H2}=\frac{1,12}{22,4}=0,05\left(mol\right)\)
\(n_{Fe}=n_{H2}=0,05\left(mol\right)\)
\(m=m_{Fe}=0,05.56=2,8\left(g\right)\)
Câu 2
\(2Al+6HCl--.2AlCl3+3H2\)
\(n_{H2}=\frac{0,336}{22,4}=0,015\left(mol\right)\)
\(n_{Al}=\frac{2}{3}n_{H2}=0,01\left(mol\right)\)
\(m_{Al}=0,01.27=0,27\left(g\right)\)
\(m_{Cu}=0,6-0,27=0,33\left(g\right)\)
Duong Le: Tức là đề bài dù là chất dư vẫn làm như k dư ạ ???
a)4 Al+3H2SO4---->2Al2(SO4)3+3H2
n\(_{Al}=\frac{5,4}{27}=0,2\left(mol\right)\)
Theo pthh
n\(_{H2}=\frac{3}{4}n_{Al}=0,15\left(mol\right)\)
V\(_{H2}=0,15.22,4=3,36\left(l\right)\)
Theo pthh
n\(_{H2SO4}=\frac{3}{4}n_{Al}=0,15\left(mol\right)\)
m\(_{H2SO4}=0,15.98=14,7\left(g\right)\)
c) H2+ CuO ---->Cu+H2O
n\(_{C_{ }uO}=\frac{4}{80}=0,05\left(mol\right)\)
=> H2 dư
Theo pthh
n\(_{Cu}=n_{CuO}=0,05\left(mol\right)\)
m\(_{Cu}=0,05.64=3,2\left(g\right)\)
\(nAl=0,2\left(mol\right)\)
\(PTHH:4Al+3H2SO4\rightarrow2Al2\left(SO4\right)3+3H2\)
=>nH2 = 3/4 Al = 0,15 (mol)
=>VH2 = 0,15 . 22,4 = 3,36l
Ta cs : nH2SO4 = 0,14 (mol)=>mH2SO4 = 0,14 . 98 = 14,7l
\(PTHH:CuO+H2\rightarrow Cu+H2O\)
\(nCuO=0,0,5\left(mol\right)\Rightarrow nCu=nCuO=0,05\left(mol\right)\)
\(\Rightarrow mCu=0,05.64=3,2\left(g\right)\)