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\(a.2Al+3H_2SO_4->Al_2\left(SO_4\right)_3+3H_2\\ n_{H_2SO_4}=\dfrac{100.29,4}{98.100}=0,3mol\\ m_{Al}=\dfrac{2}{3}.27.0,3=5,4g=m\\ V_{H_2}=22,4.0,3=6,72L\\ m_{ddsau}=m+100-0,3.2=104,8g\\b. C\%=\dfrac{342.0,1}{104,8}.100\%=32,63\%\)
a)
PTHH: 2Al+3H\(_2\)SO\(_4\) → Al\(_2\)(SO4)\(_3\)+3H\(_2\)
n\(_{H_2SO_4}\) = \(\dfrac{100.29,4\%}{98}\) = 0,3 (mol)
Theo PT: n\(_{H_2}\)= n\(_{H_2SO_4}\)= 0,3 (mol)
\(\Rightarrow\)V\(_{H_2}\)= 0,3.22,4 = 6,72(l)
Theo PT: n\(_{Al}\) = \(\dfrac{2}{3}\)n\(_{H_2}\) = 0,2 (mol)
\(\Rightarrow\)m=m\(_{Al}\) = 0,2.27 = 5,4 (g)
b)
m\(_{dd}\) = 5,4+100−0,3.2 = 104,8 (g)
Theo PT: n\(_{Al_2\left(SO_4\right)_3}\) = \(\dfrac{1}{2}\)n\(_{Al}\) = 0,1 (mol)
\(\Rightarrow\) C%\(_{Al_2\left(SO_4\right)_3}\) = \(\dfrac{0,1.342}{104,8}\).100% = 32,63%
a) $Fe + 2HCl \to FeCl_2 + H_2$
b)
n Fe = 8,4/56 = 0,15(mol) ; n HCl = 0,15.2,4 = 0,36(mol)
Ta thấy :
n Fe / 1 < n HCl /2 nên HCl dư
Theo PTHH : n H2 = n Fe = 0,15 mol
=> V = 0,15.22,4 = 3,36 lít
c) Dung dịch chứa HCl,FeCl2
m dd HCl = D.V = 0,8.150 = 120(gam)
Sau phản ứng :
n HCl dư = 0,36 - 0,15.2 = 0,06(mol)
n FeCl2 = n Fe = 0,15(mol)
m dd = 8,4 + 120 -0,15.2 = 128,1(gam)
C% HCl = 0,06.36,5/128,1 .100% = 1,71%
C% FeCl2 = 0,15.127/128,1 .100% = 14,87%
a) pthh: h2so4+ al = al2(so4)3 + h2
mH2SO4= 29,4%.100: 100%= 29,4(g)
nH2SO4= 29,4:98=0,3 mol
Ta có: nh2so4=nh2=0,3
=> Vh2=22,4.0,3=6,72 l
nH2SO4=2/3nAl=>nAl=0,2 mol
=> mAl=0,2.27=5,4g
b) mdd sau pư= mAl+ mdd H2SO4=5,4+100=105,4 g
C% H2SO4= 0,3. 98 : 105,4 .100%=30%
PTHH: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\)
Ta có: \(\left\{{}\begin{matrix}n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\n_{H_2SO_4}=\dfrac{200\cdot29,4\%}{98}=0,6\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,2}{2}< \dfrac{0,6}{3}\) \(\Rightarrow\) Axit còn dư, Nhôm p/ứ hết
\(\Rightarrow\left\{{}\begin{matrix}n_{Al_2\left(SO_4\right)_3}=0,1\left(mol\right)\\n_{H_2}=0,3\left(mol\right)\\n_{H_2SO_4\left(dư\right)}=0,6-0,3=0,3\left(mol\right)\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}m_{Al_2\left(SO_4\right)_3}=0,1\cdot342=34,2\left(g\right)\\m_{H_2SO_4\left(dư\right)}=0,3\cdot98=29,4\left(g\right)\\m_{H_2}=0,3\cdot2=0,6\left(g\right)\\V_{H_2}=0,3\cdot22,4=6,72 \left(l\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd\left(saup/ứ\right)}=m_{Al}+m_{ddH_2SO_4}-m_{H_2}=204,8\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{Al_2\left(SO_4\right)_3}=\dfrac{34,2}{204,8}\cdot100\%\approx16,7\%\\C\%_{H_2SO_4\left(dư\right)}=\dfrac{29,4}{204,8}\cdot100\%\approx14,36\%\end{matrix}\right.\)
mFe3O4 = 34.8 / 232 = 0.15 (mol)
nH2 = 8.96 / 22.4 = 0.4 (mol)
Fe3O4 + 4H2 -t0-> 3Fe + 4H2O
Bđ: 0.15.......0.4
Pư: 0.1..........0.4........0.3........0.4
Kt: 0.05.........0............0.3.......0.4
mFe3O4(dư) = 0.05 * 232 = 11.6 (g)
mFe = 0.3 * 56 = 16.8 (g)
VH2O = 0.4 * 22.4 = 8.96 (l)
2Fe + 6H2SO4(đ) => Fe2(SO4)3 + 3SO2 + 6H2O
0.3...........0.9.................0.15...........0.45
mH2SO4 = 0.9 * 98 = 88.2 (g)
C% H2SO4 = 88.2 * 100 / 98 = 90 %
VSO2 = 0.45 * 22.4 = 10.08 (l)
mX = 16.8 + 98 - 0.45 * 64 = 86 (g)
C% Fe2(SO4)3 = 0.15 * 400 / 86 * 100% = 69.76%
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%
a) pthh: H2So4 + Al -> Al2(so4)3 + H2
mH2SO4= 29,4%.100: 100%= 29,4(g)
nH2SO4= 29,4:98=0,3 mol
Ta có: nH2so4=nH2= 0,3
=> Vh2= 22,4.0,3= 6,72 l
nH2SO4= 2/3 nAl=>nAl=0,2 mol
=> mAl=0,2.27=5,4g
b) mdd sau pư= mAl+ mdd H2SO4=5,4+100=105,4 g
C% H2SO4= 0,3. 98 : 105,4 .100%=30%