Cho 2,7g nhôm(Al) tác dụng với 200g dd H2SO4, thu được muối và H2 A. Viết PTPU B. Tính VH2(đktc) C. Tính C% muôi
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 3 :
\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)
Pt : \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2|\)
2 3 1 3
0,1 0,15 0,05 0,15
a) \(n_{H2}=\dfrac{0,1.3}{2}=0,15\left(mol\right)\)
\(m_{H2}=0,15.2=0,3\left(g\right)\)
\(V_{H2\left(dktc\right)}=0,15.22,4=3,36\left(l\right)\)
b) \(n_{H2SO4}=\dfrac{0,1.3}{2}=0,15\left(mol\right)\)
⇒ \(m=0,15.98=14,7\left(g\right)\)
\(C_{ddH2SO4}=\dfrac{14,7.100}{200}=7,35\)0/0
c) \(n_{Al2\left(SO4\right)3}=\dfrac{0,15.1}{3}=0,05\left(mol\right)\)
⇒ \(m_{Al2\left(SO4\right)3}=0,05.342=17,1\left(g\right)\)
\(m_{ddspu}=2,7+200-0,3=302,4\left(g\right)\)
\(C_{Al2\left(SO4\right)3}=\dfrac{17,1.100}{302,4}=5,65\)0/0
Chúc bạn học tốt
Mình xin lỗi bạn nhé , bạn sửa lại giúp mình :
\(m_{ddspu}=2,7+200-0,3=202,4\left(g\right)\)
\(C_{Al2\left(SO4\right)3}=\dfrac{17,1.100}{202,4}=8,45\)0/0
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 2 :
\(Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\)
\(CuO+H_2\underrightarrow{^{t^0}}Cu+H_2O\)
\(K+H_2O\rightarrow KOH+\dfrac{1}{2}H_2\)
\(4P+5O_2\underrightarrow{^{ }t^0}2P_2O_5\)
\(3Fe+2O_2\underrightarrow{^{t^0}}Fe_3O_4\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
Câu 3 :
\(n_{H_2}=\dfrac{2.7}{27}=0.1\left(mol\right)\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
\(0.1...........................0.05.......0.15\)
\(V_{H_2}=0.15\cdot22.4=3.36\left(l\right)\)
\(m_{Al_2\left(SO_4\right)_3}=0.05\cdot342=17.1\left(g\right)\)
\(m_{\text{dung dịch sau phản ứng}}=2.7+200-0.15\cdot2=202.4\left(g\right)\)
\(C\%_{Al_2\left(SO_4\right)_3}=\dfrac{17.1}{202.4}\cdot100\%=8.44\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a,\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)
PTHH: 2Al + 3H2SO4 → Al2(SO4)3 + 3H2
Mol: 0,1 0,15 0,05 0,15
\(m_{H_2}=0,15.2=0,3\left(g\right)\)
\(V_{H_2}=0,15.24,79=3,7185\left(l\right)\)
b, \(m_{ddH_2SO_4}=\dfrac{0,15.98.100\%}{200}=7,35\%\)
c, mdd sau pứ = 2,7 + 200 - 0,3 = 202,4 (g)
\(C\%_{ddAl_2\left(SO_4\right)_3}=\dfrac{0,05.342.100\%}{202,4}=8,45\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{H_2}=\dfrac{2.7}{27}=0.1\left(mol\right)\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
\(0.1..........................0.05............0.15\)
\(V_{H_2}=0.15\cdot22.4=3.36\left(l\right)\)
\(m_{Al_2\left(SO_4\right)_3}=0.05\cdot342=17.1\left(g\right)\)
\(m_{\text{dung dịch sau phản ứng}}=2.7+200-0.15\cdot2=202.4\left(g\right)\)
\(C\%_{Al_2\left(SO_4\right)_3}=\dfrac{17.1}{202.4}\cdot100\%=8.45\%\)
Ta có: \(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)
a, PT: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
_____0,1____0,15_______0,05_____0,15 (mol)
b, Ta có: \(V_{H_2}=0,15.22,4=3,36\left(l\right)\)
c, Ta có: m dd sau pư = mAl + m dd H2SO4 - mH2 = 2,7 + 200 - 0,15.2 = 202,4 (g)
\(\Rightarrow C\%_{Al_2\left(SO_4\right)_3}=\dfrac{0,05.342}{202,4}.100\%\approx8,45\%\)
Bạn tham khảo nhé!
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
\(n_{H_2SO_4}=\dfrac{300.19,6\%}{98}=0,6\left(mol\right);n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\\ PTHH:2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ Vì:\dfrac{0,1}{2}< \dfrac{0,6}{3}\Rightarrow H_2SO_4dư\\ n_{Al_2\left(SO_4\right)_3}=\dfrac{3n_{Al}}{2}=\dfrac{3.0,1}{2}=0,15\left(mol\right)\\ a,m_{Al_2\left(SO_4\right)_3}=342.0,15=51,3\left(g\right)\\ b,m_{ddsau}=m_{Al}+m_{ddH_2SO_4}-m_{H_2}=2,7+300-\dfrac{3}{2}.0,1.2=302,4\left(g\right)\\ c,C\%_{ddAl_2\left(SO_4\right)_3}=\dfrac{51,3}{302,4}.100\%\approx16,964\%\\ n_{H_2SO_4\left(dư\right)}=0,6-\dfrac{3}{2}.0,1=0,45\left(mol\right)\\ C\%_{ddH_2SO_4\left(dư\right)}=\dfrac{0,45.98}{302,4}.100\%\approx14,583\%\)
Bài 2:
\(PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ Đặt:n_{Al}=a\left(mol\right);n_{Mg}=b\left(mol\right)\left(a,b>0\right)\\ Hpt:\left\{{}\begin{matrix}27a+24b=7,8\\1,5a+b=\dfrac{8,96}{22,4}=0,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\\ \%m_{Al}=\dfrac{0,2.27}{7,8}.100\%\approx69,231\%\Rightarrow\%m_{Mg}\approx100\%-69,231\%\approx30,769\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Sửa: \(14,7\%\)
\(n_{H_2SO_4}=\dfrac{200.14,7\%}{100\%.98}=0,3(mol)\\ a,PTHH:2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\\ \Rightarrow n_{H_2}=0,3(mol)\\ \Rightarrow V_{H_2}=0,3.22,4=6,72(l)\\ b,n_{Al}=\dfrac{2}{3}n_{H_2SO_4}=0,2(mol)\\ \Rightarrow m_{Al}=0,2.27=5,4(g)\\ c,n_{Al_2(SO_4)_3}=\dfrac{1}{2}n_{Al}=0,1(mol)\\ \Rightarrow C\%_{Al_2(SO_4)_3}=\dfrac{0,1.342}{200+5,4-0,3.2}.100\%=16,7\%\\ c,m_{Al_2(SO_4)_3}=0,1.342=34,2(g)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right);n_{H_2SO_4}=\dfrac{245.20\%}{98}=0,5\left(mol\right)\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
0,2..........0,5
Lập tỉ lệ : \(\dfrac{0,2}{2}< \dfrac{0,5}{3}\)
=> H2SO4 dư
\(n_{H_2}=\dfrac{3}{2}n_{Al}=0,3\left(mol\right)\)
=> V H2 = 0,3.22,4= 6,72(l)
\(m_{ddsaupu}=5,4+245-0,3.2=249,8\left(g\right)\)
=> \(C\%_{Al_2\left(SO_4\right)_3}=\dfrac{0,1.342}{249,8}.100=13,69\%\)
a) mH2SO4=20%.245=49(g) ->nH2SO4=49/98=0,5(mol)
nAl=5,4/27=0,2(mol)
PTHH: 2Al +3 H2SO4 -> Al2(SO4)3 +3 H2
Ta có: 0,2/2 < 0,5/3
=> H2SO4 dư, Al hết, tính theo nAl
=> nH2SO4(p.ứ)=nH2=3/2. nAl=3/2. 0,2= 0,3(mol)
=> nH2SO4(dư)=0,5 - 0,3=0,2(mol)
=>mH2SO4(dư)=0,2.98=19,6(g)
b) V(H2,đktc)=0,3.22,4=6,72(l)
c) nAl2(SO4)3= 1/2. nAl=1/2. 0,2=0,1(mol)
=>mAl2(SO4)3=342.0,1=34,2(g)
mddAl2(SO4)3=mAl+ mddH2SO4-mH2=5,4+245 - 0,3.2= 249,8(g)
=>C%ddAl2(SO4)3= (34,2/249,8).100=13,691%
2Al + 3H2SO4 -> Al2(SO4)3 + 3H2
nAl = 0,1 => nH2 = 0,15 => VH2 = 0,15 . 22,4 = 3,36 (l)
nH2 = 0,15 => mH2 = 0,3(g)
m dd sau pư = 2,7 + 200 -0,3=202,4 (g)
theo pư => n Al2(SO4)3 = 0,05 => m Al2(SO4)3 = 17,1 => C% = 17,1:202,4 . 100 % = 8,45%
A)
$2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2$
B)
n Al = 2,7/27 = 0,1(mol)
Theo PTHH :
n H2 = 3/2 n Al = 0,15(mol)
=> V H2 = 0,15.22,4 = 3,36(lít)
Theo PTHH :
n Al2(SO4)3 = 1/2 n Al = 0,05(mol)
m dd sau pư = m Al + mdd H2SO4 - m H2 = 2,7 + 200 - 0,15.2 = 202,4 gam
Suy ra :
C% Al2(SO4)3 = 0,05.342/202,4 .100% = 8,45%