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$a) 2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2$
$b) n_{Al} = \dfrac{10,8}{27} = 0,4(mol)$
Theo PTHH : $n_{H_2SO_4} = \dfrac{3}{2}n_{Al} = 0,6(mol)$
$m_{H_2SO_4} = 0,6.98 = 58,8(gam)$
$c) n_{Al_2(SO_4)_3} = \dfrac{1}{2}n_{Al} = 0,2(mol) \Rightarrow m_{Al_2(SO_4)_3} = 0,2.342 = 68,4(gam)$
$d) n_{H_2} = n_{H_2SO_4} = 0,6(mol) \Rightarrow V_{H_2} = 0,6.22,4 = 13,44(lít)$
\(a,2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
\(b.n_{Al}=\dfrac{m}{M}=0,4\left(mol\right)\)
\(Theo.PTHH\Rightarrow n_{H_2SO_4}=n_{H_2}=\dfrac{3}{2}n_{Al}=1,5.0,4=0,6\left(mol\right)\\ \Rightarrow m_{H_2SO_4}=n.M=0,6.98=58,8\left(g\right)\)
\(c,Theo.PTHH\Rightarrow n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Al}=0,5.0,4=0,2\left(mol\right)\\ \Rightarrow m_{Al_2\left(SO_4\right)_3}=n.M=0,2.342=68,4\left(g\right)\\ d,V_{H_2\left(dktc\right)}=n.22,4=0,6.22,4=13,44\left(l\right)\)
\(n_{Al}=\dfrac{1,35}{27}=0,05\left(mol\right)\\ 2Al+3H_2SO_4\rightarrow2Al_2\left(SO_4\right)_3+3H_2\\ n_{H_2SO_4}=n_{H_2}=\dfrac{3}{2}.0,05=0,075\left(mol\right)\\ n_{Al_2\left(SO_4\right)_3}=\dfrac{0,05}{2}=0,025\left(mol\right)\\ a,m_{Al_2\left(SO_4\right)_3}=342.0,025=8,55\left(g\right)\\ b,V_{H_2\left(đktc\right)}=0,075.22,4=1,68\left(l\right)\\ c,m_{H_2SO_4}=0,075.98=7,35\left(g\right)\)
\(n_{Al}=\dfrac{1,35}{27}=0,05mol\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
0,05 0,075 0,025 0,075
\(m_{Al_2\left(SO_4\right)_3}=0,025\cdot342=8,55g\)
\(V_{H_2}=0,075\cdot22,4=1,68l\)
\(m_{H_2SO_4}=0,075\cdot98=7,35g\)
a) \(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\)
PTHH: 2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
0,4-->0,6---------->0,2------->0,6
=> \(C_{M\left(dd.H_2SO_4\right)}=\dfrac{0,6}{0,15}=4M\)
b) VH2 = 0,6.22,4 = 13,44 (l)
c) \(C_{M\left(Al_2\left(SO_4\right)_3\right)}=\dfrac{0,2}{0,15}=\dfrac{4}{3}M\)
\(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\\
pthh:2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
0,4 0,6 0,2 0,6
\(C_M_{H_2SO_4}=\dfrac{0,6}{0,15}=4M\\ V_{H_2}=0,622,4=13,44L\)
\(C_M=\dfrac{0,2}{0,15}=1,3M\)
\(n_{Al}=\dfrac{10.8}{27}=0.4\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{22.4}{98}=\dfrac{8}{35}\left(mol\right)\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
\(2............3\)
\(0.4..........\dfrac{8}{35}\)
\(LTL:\dfrac{0.4}{2}>\dfrac{\dfrac{8}{35}}{3}\Rightarrow Aldư\)
\(m_{Al\left(dư\right)}=\left(0.4-\dfrac{8}{35}\cdot\dfrac{2}{3}\right)\cdot27=6.68\left(g\right)\)
\(m_{Al_2\left(SO_4\right)_3}=\dfrac{8}{35\cdot3}\cdot342=26.05\left(g\right)\)
\(V_{H_2}=\dfrac{8}{35}\cdot22.4=5.12\left(l\right)\)
1. Gọi mol của Mg và Al là x, y mol
=> 24x + 27y = 12,6 (1)
nH2 = 0,6 mol => x + 1,5y = 0,6 (2)
Từ (1) (2) => x = 0,3 ; y = 0,2
=> %Mg = 57,14%
=> %Al = 42,86%
1. Gọi mol của Mg và Al là x, y mol
=> 24x + 27y = 12,6 (1)
nH2 = 0,6 mol => x + 1,5y = 0,6 (2)
Từ (1) (2) => x = 0,3 ; y = 0,2
=> %Mg = 57,14%
=> %Al = 42,86%
1. Gọi mol của Mg và Al là x, y mol
=> 24x + 27y = 12,6 (1)
nH2 = 0,6 mol => x + 1,5y = 0,6 (2)
Từ (1) (2) => x = 0,3 ; y = 0,2
=> %Mg = 57,14%
=> %Al = 42,86%
Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
PT: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
Theo PT: \(n_{H_2SO_4}=n_{FeSO_4}=n_{H_2}=n_{Fe}=0,1\left(mol\right)\)
a, \(V_{H_2}=0,1.24,79=2,479\left(l\right)\)
b, \(m_{H_2SO_4}=0,1.98=9,8\left(g\right)\)
\(\Rightarrow m_{ddH_2SO_4}=\dfrac{9,8}{9,8\%}=100\left(g\right)\)
c, Ta có: m dd sau pư = 5,6 + 100 - 0,1.2 = 105,4 (g)
\(\Rightarrow C\%_{FeSO_4}=\dfrac{0,1.152}{105,4}.100\%\approx14,42\%\)
nAl= 0,04(mol)
PTHH: 2 Al + 3 H2SO4 -> Al2(SO4)3 + 3 H2
0,04___________0,06___0,02_____0,06(mol)
a) V(H2, đktc)=0,06.22,4=1,344(l)
b) VddH2SO4= 0,06/2=0,03(l)=30(ml)
c) VddAl2(SO4)3=VddH2SO4=0,03(l)
=>CMddAl2(SO4)3=0,02/0,03=2/3(M)
\(n_{Al}=\dfrac{1.08}{27}=0.04\left(mol\right)\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
\(0.04......0.06.............0.02...........0.06\)
\(V_{H_2}=0.06\cdot22.4=1.344\left(l\right)\)
\(V_{dd_{H_2SO_4}}=\dfrac{0.06}{2}=0.03\left(l\right)\)
\(C_{M_{Al_2\left(SO_4\right)_3}}=\dfrac{0.02}{0.03}=\dfrac{2}{3}\left(M\right)\)
Tại sao mdd bằng 244,8 gam vậy em?
2Al+3H2SO4->Al2(SO4)3+3H2
0,2------0,3--------------------- 0,3
nAl=5,4\27=0,2 mol
=>=>VH2=0,3.22,4=6,72l
=>mH2SO4=0,3.98=29,4g
=>nAl2(SO4)3 = 0,1 (mol)
=>mAl2(SO4)3 = 0,1 . 342 = 34 , 2(g)
Ta có : mH2 = 0,6 (mol)
=>mdd =244,8 (g)
=>C% = 13,97%