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2Al+3H2SO4->Al2(SO4)3+3H2
0,1-------0,15---------------------0,15 mol
n H2=\(\dfrac{3,36}{22,4}\)=0,15 mol
=>m Al=0,1.27=2,7g
=>m H2SO4=0,15.98=14,7g
a, PT: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
b, Ta có: \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
Theo PT: \(n_{Al}=\dfrac{2}{3}n_{H_2}=0,1\left(mol\right)\)
\(\Rightarrow m_{Al}=0,1.27=2,7\left(g\right)\)
c, Theo PT: \(n_{H_2SO_4}=n_{H_2}=0,15\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}=0,15.98=14,7\left(g\right)\)
Bạn tham khảo nhé!
\(n_{Fe}=\dfrac{22,4}{56}=0,4\left(mol\right)\\ a,Fe+2HCl\rightarrow FeCl_2+H_2\\ b,n_{HCl}=0,4.2=0,8\left(mol\right)\\ m_{HCl}=0.8.36,5=29,2\left(g\right)\\ c,n_{H_2}=n_{FeCl_2}=n_{Fe}=0,4\left(mol\right)\\ m_{FeCl_2}=0,4.127=50,8\left(g\right)\\ d,V_{H_2\left(dktc\right)}=0,4.22,4=8,96\left(l\right)\)
a)
\(2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2\)
b)
\(n_{H_2} = \dfrac{6.13,44}{22,4} = 3,6(mol)\)
Theo PTHH :
\(n_{Al} = \dfrac{2}{3}n_{H_2} = 2,4(mol)\\ \Rightarrow m_{Al} = 2,4.27 = 64,8(gam)\)
c)
\(4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3\)
Theo PT trên :
\(n_{O_2} = \dfrac{3}{4}n_{Al} = 1,8(mol)\\ \Rightarrow V_{O_2} = 1,8.22,4 = 40,32(lít)\)
\(n_{Al}=\dfrac{10.8}{27}=0.4\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{22.4}{98}=\dfrac{8}{35}\left(mol\right)\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
\(2............3\)
\(0.4..........\dfrac{8}{35}\)
\(LTL:\dfrac{0.4}{2}>\dfrac{\dfrac{8}{35}}{3}\Rightarrow Aldư\)
\(m_{Al\left(dư\right)}=\left(0.4-\dfrac{8}{35}\cdot\dfrac{2}{3}\right)\cdot27=6.68\left(g\right)\)
\(m_{Al_2\left(SO_4\right)_3}=\dfrac{8}{35\cdot3}\cdot342=26.05\left(g\right)\)
\(V_{H_2}=\dfrac{8}{35}\cdot22.4=5.12\left(l\right)\)
2Al+3H2SO4->al2(SO4)3+3H2
Fe+H2SO4->FeSO4+H2
Gọi x,y tương ứng là số mol của Al và Fe:
Ta có: 27x+56y=11 (1)
nH2=0,4 mol
1,5x+y=0,4 (2)
Giải hệ(1),(2):x=0,2;y=0,1
mAl=0,2.27=5,4g
%Al=\(\dfrac{5,4.100}{16,6}\)=32,53%
=>%Fe=67,47%
m H2SO4=0,4.98=39,2g
c) m muối=0,1.342+0,1.152=49,4g
$a) 2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2$
$b) n_{Al} = \dfrac{10,8}{27} = 0,4(mol)$
Theo PTHH : $n_{H_2SO_4} = \dfrac{3}{2}n_{Al} = 0,6(mol)$
$m_{H_2SO_4} = 0,6.98 = 58,8(gam)$
$c) n_{Al_2(SO_4)_3} = \dfrac{1}{2}n_{Al} = 0,2(mol) \Rightarrow m_{Al_2(SO_4)_3} = 0,2.342 = 68,4(gam)$
$d) n_{H_2} = n_{H_2SO_4} = 0,6(mol) \Rightarrow V_{H_2} = 0,6.22,4 = 13,44(lít)$
\(a,2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
\(b.n_{Al}=\dfrac{m}{M}=0,4\left(mol\right)\)
\(Theo.PTHH\Rightarrow n_{H_2SO_4}=n_{H_2}=\dfrac{3}{2}n_{Al}=1,5.0,4=0,6\left(mol\right)\\ \Rightarrow m_{H_2SO_4}=n.M=0,6.98=58,8\left(g\right)\)
\(c,Theo.PTHH\Rightarrow n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Al}=0,5.0,4=0,2\left(mol\right)\\ \Rightarrow m_{Al_2\left(SO_4\right)_3}=n.M=0,2.342=68,4\left(g\right)\\ d,V_{H_2\left(dktc\right)}=n.22,4=0,6.22,4=13,44\left(l\right)\)