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nH2SO4 = 14,7: 27=0,54(mol)
PTHH : 2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
theo pt , nH2 = nH2SO4=0,54(mol)
=> VH2(đktc) = 0,54. 22,4=12,096 (l)
b theo pt nAl = 3/2. nH2=0,36 (mol)
=> mAl = 0,36.27 =9,72(g)
c)theo pt n Al2(SO4)3 = 1/2nAl = 0,18(mol)
=>mAl2(SO4)3= 0,18.342=61,56(g)
\(a,PTHH:2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ \Rightarrow n_{Al}=\dfrac{2}{3}n_{H_2}=0,1\left(mol\right)\\ \Rightarrow m_{Al}=0,1\cdot27=2,7\left(g\right)\\ b,n_{H_2SO_4}=n_{H_2}=0,15\left(mol\right)\\ \Rightarrow m_{H_2SO_4}=0,15\cdot98=14,7\left(g\right)\\ c,n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{3}n_{H_2}=0,05\left(mol\right)\\ \Rightarrow m_{Al_2\left(SO_4\right)_3}=0,05\cdot342=17,1\left(g\right)\)
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!
a. 2Al + 6HCl -> 2AlCl3 + 3H2
b. nAl = \(\dfrac{8.1}{27}=0,3\left(mol\right)\)=> \(n_{H_2}=\dfrac{3}{2}.0,3=0,45\left(mol\right)\)
\(V_{H_2}=0,45.22,4=10,08\left(mol\right)\)
a. PTHH:
Al+H2SO4-->AlSO4+H2
b.Theo ĐLBTKL, ta có:
mAl+mH2SO4=mAl2SO4+mH2
=>mH2SO4=mAl2SO4+mH2-mAl
=171+3-27=147 (g)
nH2= 13,44 : 22,4 = 0,6 (mol)
pthh : 2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
0,4 <------------------- 0,2<-----------<0,6 (mol)
mAl = 0,4 . 27 = 10,8 ( g)
mAl2(SO4)3= 0,2 . 342 = 68,4 (g)
nFe3O4 = 46,4 : 232 = 0,2 (mol)
pthh : Fe3O4 + 4H2 -t--> 3Fe + 4H2O
LTL :
0,2/1 > 0,6 /4
=> Fe3O4 du
theo pt nFe=3/4 nH2 = ,45 (mol)
=> mFe= 0,45 . 56= 25,2 (g)
\(a,n_{H_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\\ PTHH:2Al+3H_2SO_4\rightarrow2Al_2\left(SO_4\right)_3+3H_2\uparrow\\ Theo.pt:n_{Al}=n_{Al_2\left(SO_4\right)_3}=\dfrac{2}{3}n_{H_2SO_4}=\dfrac{2}{3}.0,6=0,4\left(mol\right)\\ m_{Al}=0,4.27=10,8\left(g\right)\\ b,m_{Al_2\left(SO_4\right)_3}=0,4.342=136,8\left(g\right)\\ c,n_{Fe_3O_4}=\dfrac{46,4}{232}=0,2\left(mol\right)\\ PTHH:Fe_3O_4+4H_2\underrightarrow{t^o}3Fe+4H_2O\\ LTL:\dfrac{0,2}{1}>\dfrac{0,6}{4}\Rightarrow Fe_3O_4.du\\ n_{Fe}=\dfrac{3}{4}n_{H_2}=\dfrac{3}{4}.0,4=0,3\left(mol\right)\\ m_{Fe}=0,3.56=16,8\left(g\right)\)