\(n_{Fe_2O_3}=\dfrac{3.2}{160}=0.02\left(mol\right)\)

\(n_{HCl}=\dfrac{2.19}{36.5}=0.06\left(mol\right)\)

\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)

\(1...........6\)

\(0.02...........0.06\)

Lập tỉ lệ : \(\dfrac{0.02}{1}>\dfrac{0.06}{6}\Rightarrow Fe_2O_3dư\)

\(n_{Fe_2O_3\left(dư\right)}=0.02-\dfrac{0.06}{6}=0.01\left(mol\right)\)

\(m_{Fe_2O_3\left(dư\right)}=0.01\cdot160=1.6\left(g\right)\)

\(m_{FeCl_3}=0.02\cdot162.5=3.25\left(g\right)\)

\(m_{H_2O}=0.03\cdot18=0.54\left(g\right)\)

12,395l ở đk nào vậy em?

pthhL

\(3O_2+4Al\underrightarrow{t^o}2Al_2O_3\)

\(nO_2=12,395:24,9=0,5\left(mol\right)\)

\(nAl=\dfrac{10,8}{27}=0,4\left(mol\right)\)

Xét tỉ lê:

\(\dfrac{nO_2}{3}>\dfrac{nAl}{4}\)

=> Oxy dư và dư: \(mO_{2dư}=0,06.32=2g\)

b, tính theo pthh vì nO2 dư nên lấy số mol của Al

=> \(nAl_2O_3=\dfrac{2}{4}.0,4=0,2\left(mol\right)\)

\(\Rightarrow mAl_2O_3=0,2.102=20,4\left(g\right)\)

đktc ạ 

Ta có: \(n_{NaOH}=\dfrac{40.10\%}{40}=0,1\left(mol\right)\)

\(n_{HCl}=\dfrac{100.14,6\%}{36,5}=0,4\left(mol\right)\)

a, PT: \(NaOH+HCl\rightarrow NaCl+H_2O\)

Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,4}{1}\), ta được HCl dư.

Theo PT: \(n_{HCl\left(pư\right)}=n_{NaCl}=n_{NaOH}=0,1\left(mol\right)\)

\(\Rightarrow n_{HCl\left(dư\right)}=0,4-0,1=0,3\left(mol\right)\)

\(\Rightarrow m_{HCl\left(dư\right)}=0,3.36,5=10,95\left(g\right)\)

b, \(C\%_{HCl}=\dfrac{10,95}{40+100}.100\%\approx7,82\%\)

\(C\%_{NaCl}=\dfrac{0,1.58,5}{40+100}.100\%\approx4,18\%\)

\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ n_{HCl}=\dfrac{25,55}{36,5}=0,7\left(mol\right)\\a. 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ b.Vì:\dfrac{0,2}{2}< \dfrac{0,7}{6}\\ \Rightarrow HCldư\\ n_{HCl\left(dư\right)}=0,7-\dfrac{6}{2}.0,2=0,1\left(mol\right)\\ n_{AlCl_3}=n_{Al}=0,2\left(mol\right)\\ n_{H_2}=\dfrac{3}{2}.0,2=0,3\left(mol\right)\\ m_{H_2}=0,3.2=0,6\left(g\right)\\ m_{HCl\left(dư\right)}=0,1.36,5=3,65\left(g\right)\\ m_{AlCl_3}=133,5.0,2=26,7\left(g\right)\)

\( n_{Al}=\dfrac{2,7}{27}=0,1mol\)

\(n_{H_2SO_4}=\dfrac{24,5}{98}=0,25mol\)

\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

0,1        0,25           0                    0

0,1        0,15           0,05               0,15

0           0,1             0,05               0,15

Chất \(H_2SO_4\) dư và dư \(m=0,1\cdot98=9,8g\)

\(V_{H_2}=0,15\cdot22,4=3,36l\)

\(nAl=\dfrac{2,7}{27}=0,1\left(mol\right)\)

\(nH_2SO_4=\dfrac{24,5}{98}=0,25\left(mol\right)\)

\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\)

0,1---->0,15------>0,05--------------->0,15

Xét tỉ lệ : \(\dfrac{0,1}{2}< \dfrac{0,25}{3}\)

=> H2SO4 dư vs pứ

\(nH_2SO_{4\left(dư\right)}=0,25-0,15=0,1\left(mol\right)\)

\(mH_2SO_4=\)\(0,1.98=9,8\left(g\right)\)

\(VH_2=0,15.22,4=3,36\left(lít\right)\)

a) Fe₂O₃ + 6HCl --> 2FeCl₃ + 3H₂O

\(n_{Fe_2O_3}=\dfrac{3,2}{160}=0,02\left(mol\right)\)

\(n_{HCl}=\dfrac{2,19}{36,5}=0,06\left(mol\right)\)

Xét tỉ lệ \(\dfrac{0,02}{1}>\dfrac{0,06}{6}\) => Fe₂O₃ dư, HCl hết

PTHH: Fe₂O₃ + 6HCl --> 2FeCl₃ + 3H₂O

            0,01<--0,06------->0,02---->0,03

=> \(m_{Fe_2O_3\left(dư\right)}=\left(0,02-0,01\right).160=1,6\left(g\right)\)

b) \(m_{FeCl_3}=0,02.162,5=3,25\left(g\right)\)

\(m_{H_2O}=0,03.18=0,54\left(g\right)\)

nFe = 2.8/56 = 0.05 (mol) 

nHCl = 14.6/36.5 = 0.4 (mol) 

Fe + 2HCl => FeCl2 + H2 

1.........2

0.05......0.4 

LTL : 0.05/1 < 0.4/2 

=> HCl dư 

mHCl (dư) = ( 0.4 - 0.1 ) * 36.5 = 10.95 (g)

VH2 = 0.05*22.4 = 1.12 (l)

nHCl (dư) = 0.4 - 0.1 = 0.3 (mol)  

mFe cần thêm = 0.3/2 * 56 = 8.4 (g) 

\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right);n_{HCl}=0,5.1=0,5\left(mol\right)\\ PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ Vì:\dfrac{0,5}{2}>\dfrac{0,1}{1}\Rightarrow Zn.hết,HCldư\\ n_{HCl\left(dư\right)}=0,5-2.0,1=0,3\left(mol\right)\\ m_{HCl\left(dư\right)}=0,3.36,5=10,95\left(g\right)\)