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\(n_{Al}=\dfrac{10.8}{27}=0.4\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{22.4}{98}=\dfrac{8}{35}\left(mol\right)\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
\(2............3\)
\(0.4..........\dfrac{8}{35}\)
\(LTL:\dfrac{0.4}{2}>\dfrac{\dfrac{8}{35}}{3}\Rightarrow Aldư\)
\(m_{Al\left(dư\right)}=\left(0.4-\dfrac{8}{35}\cdot\dfrac{2}{3}\right)\cdot27=6.68\left(g\right)\)
\(m_{Al_2\left(SO_4\right)_3}=\dfrac{8}{35\cdot3}\cdot342=26.05\left(g\right)\)
\(V_{H_2}=\dfrac{8}{35}\cdot22.4=5.12\left(l\right)\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
\(n_{H_2}=\dfrac{33,44}{22,4}=1,5mol\)
\(\Rightarrow n_{Al}=\dfrac{1,5}{3}.2=1mol\) \(\Rightarrow m_{Al}=1.27=27g\)
\(n_{H_2SO_4}=n_{H_2}=1,5mol\) \(\Rightarrow m_{H_2SO_4}=1,5.98=147g\)
\(n_{Al_2\left(SO_4\right)_3}=\dfrac{1,5}{3}=0,5mol\) \(\Rightarrow m_{Al_2\left(SO_4\right)_3}=0,5.342=171g\)
a) 3H2SO4 + 2Al ---> Al2(SO4)3 + 3H2
0,75 0,5 0,25 0,75 ( mol )
b) mAl = 0,5 . 27 =13,5 g
c) VH2= 0,75 . 22,4 = 16,8 l
d) mAl2(SO4)3 = 0,25 . 342 =85,5g
e) nAl = 0,5 mol
nH2 = 0,75 mol
a) PTHH: 2Al+ 3H2SO4 ----> Al2(SO4)3 + 3H2
0.5 0.75 0.25 0.75
b)mAl=0.5*27=13.5g
c)VH2=0.75*22.4=16.8 l
d)mAl2(SO4)3=0.25*342=85.5g
c)số mol của Al và H2 đã có trên PTHH
Chúc em học tốt!!!
nAl=8,1/27=0,3mol ;nH2SO4=53,9/98=0,55mol
ta có pt : 2Al+3H2SO4---->Al2(SO4)3+3H2
Trước p/u: 0,3mol 0,55mol
p/u : 0,3mol 0,45mol
Saup/u: 0mol 0,1mol 0,15mol 0,45mol
=>H2SO4 dư
mH2SO4 dư =0,1.98=9,8g
b,mAl2SO43 =0,15.294=44,1g
c, mH2=0,45.2=0,9g
V H2=0,45.22,4=10,08l
\(n_{Al}=\dfrac{1,35}{27}=0,05\left(mol\right)\\ 2Al+3H_2SO_4\rightarrow2Al_2\left(SO_4\right)_3+3H_2\\ n_{H_2SO_4}=n_{H_2}=\dfrac{3}{2}.0,05=0,075\left(mol\right)\\ n_{Al_2\left(SO_4\right)_3}=\dfrac{0,05}{2}=0,025\left(mol\right)\\ a,m_{Al_2\left(SO_4\right)_3}=342.0,025=8,55\left(g\right)\\ b,V_{H_2\left(đktc\right)}=0,075.22,4=1,68\left(l\right)\\ c,m_{H_2SO_4}=0,075.98=7,35\left(g\right)\)
\(n_{Al}=\dfrac{1,35}{27}=0,05mol\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
0,05 0,075 0,025 0,075
\(m_{Al_2\left(SO_4\right)_3}=0,025\cdot342=8,55g\)
\(V_{H_2}=0,075\cdot22,4=1,68l\)
\(m_{H_2SO_4}=0,075\cdot98=7,35g\)
\(n_{Al}=\dfrac{5.4}{27}=0.2\left(mol\right)\\ n_{H_2SO_4}=\dfrac{49}{98}=0.5\left(mol\right)\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
\(Bđ:0.2..........0.5\)
\(Pư:0.2.........0.3..............0.1............0.3\)
\(Kt:0...........0.2...............0.1.............0.3\)
\(m_{Al_2\left(SO_4\right)_3}=0.1\cdot342=34.2\left(g\right)\)
\(V_{H_2}0.3\cdot22.4=6.72\left(l\right)\)
\(a,PTHH:2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ \Rightarrow\text{Số nguyên tử Al là }2\\ b,n_{H_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\\ \Rightarrow n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{3}n_{H_2}=0,2\left(mol\right)\\ \Rightarrow m_{Al_2\left(SO_4\right)_3}=0,2\cdot342=68,4\left(g\right)\\ c,C_1:n_{H_2SO_4}=n_{H_2}=0,6\left(mol\right)\\ \Rightarrow m_{H_2SO_4}=0,6\cdot98=58,8\left(g\right)\\ C_2:n_{Al}=\dfrac{2}{3}n_{H_2}=0,4\left(mol\right)\\ \Rightarrow m_{Al}=0,4\cdot27=10,8\left(g\right)\\ m_{H_2}=0,6\cdot2=1,2\left(g\right)\\ \text{Bảo toàn KL: }m_{H_2SO_4}=m_{Al_2\left(SO_4\right)_3}+m_{H_2}-m_{Al}=68,4+1,2-10,8=58,8\left(g\right)\)
2Al + 3H2SO4 -> Al2(SO4)3 + 3H2
nAl=\(\dfrac{10,8}{27}=0,4\left(mol\right)\)
Theo PTHH ta có:
\(\dfrac{3}{2}\)nAl=nH2SO4=nH2=0,6(mol)
\(\dfrac{1}{2}\)nAl=nAl2(SO4)3=0,2(mol)
mH2SO4=98.0,6=58,8(g)
VH2=0,6.22,4=13,44(lít)
C1:mAl2(SO4)3=0,2.342=68,4(g)
C2:
Áp dụng định luật BTKL ta có:
mAl+mH2SO4=mAl2(SO4)3+mH2
=>mAl2(SO4)3=mAl+mH2SO4-mH2
=10,8+58,8-0,6.2=68,4(g)