10.

\(n_{Fe}=\dfrac{16.8}{56}=0.3\left(mol\right)\)

\(4Fe+3O_2\underrightarrow{^{^{t^0}}}2Fe_2O_3\)

\(0.3.....0.225....0.15\)

\(V_{O_2}=0.225\cdot22.4=5.04\left(l\right)\)

\(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)

\(0.15...........0.45\)

\(m_{H_2SO_4}=0.45\cdot98=44.1\left(g\right)\)

11.

\(n_{Fe_2O_3}=\dfrac{48}{160}=0.3\left(mol\right)\)

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

\(0.3...........1.8...........0.6\)

\(m_{FeCl_3}=0.6\cdot162.5=97.5\left(g\right)\)

\(m_{HCl}=1.8\cdot36.5=65.7\left(g\right)\)

Bài 10:

\(a,n_{Fe}=\dfrac{16,8}{56}=0,3(mol)\\ PTHH:4Fe+3O_2\xrightarrow{t^o}2Fe_2O_3\\ Fe_2O_3+3H_2SO_4\to Fe_2(SO_4)_3+3H_2O\\ \Rightarrow n_{O_2}=\dfrac{3}{4}n_{Fe}=0,225(mol)\\ \Rightarrow V_{O_2}=0,225.22,4=5,04(l)\\ b,n_{H_2SO_4}=3n_{Fe_2O_3}=3.\dfrac{1}{2}n_{Fe}=0,45(mol)\\ \Rightarrow m_{H_2SO_4}=0,45.98=44,1(g)\)

Bài 11:

\(a,n_{Fe_2O_3}=\dfrac{48}{160}=0,3(mol)\\ PTHH:Fe_2O_3+6HCl\to 2FeCl_3+3H_2O\\ \Rightarrow n_{FeCl_3}=2n_{Fe_2O_3}=0,6(mol)\\ \Rightarrow m_{FeCl_3}=0,6.162,5=97,5(g)\\ b,n_{HCl}=6n_{Fe_2O_3}=1,8(mol)\\ \Rightarrow m_{HCl}=1,8.36,5=65,7(g)\)

Em tách từng câu hỏi gồm 1 hoặc 2 bài nhé ! Đăng nhiều như thế này rối lắm.

\(4Al+3O_2\rightarrow\left(t^o\right)Al_2O_3\\ Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\\ a,n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\\ n_{O_2}=\dfrac{3}{4}.0,3=0,225\left(mol\right)\\ V_{O_2\left(\text{đ}ktc\right)}=0,225.22,4=5,04\left(l\right)\\ b,n_{Al_2O_3}=\dfrac{1}{4}.0,3=0,075\left(mol\right)\\ n_{H_2SO_4}=3.0,075=0,225\left(mol\right)\\ m_{H_2SO_4}=m=0,225.98=22,05\left(g\right)\)

a) PTHH: \(4Fe+3O_2\underrightarrow{t^o}2Fe_2O_3\) 

Ta có: \(n_{Fe}=\dfrac{m}{M}=\dfrac{16,8}{56}=0,3\left(mol\right)\)

PTPƯ: 4Fe         +         3O₂          \(\underrightarrow{t^o}\)          2Fe₂O₃

             4                         3                              2

            0,3                    0,225                         0,15

\(\Rightarrow V_{O_2}=n.24,79=0,225.24,79=5,57775\left(l\right)\) 

TT:mFe=16.8g

a/Tìm VO2?

b/Tìm mH2SO4?

Giải:a/ 

PTHH: t° 

nFe=mFe/M=16.8/56=0,3mol

Theo PTHH:4molFe-->3molO2 

0,3molFe-->nO2=0,3molFe.3/4=0,225mol

4molFE-->2molFe2O3

0,3molFe-->nFe2O3=0,3molFe.2/4=0,15mol

VO2=nO2.22,4=0,225.22,4=5,04l

b/PTHH:Fe2O3+3H2SO4->Fe2(SO4)3+3H2O

Theo PTHH:1molFe2O3-->3molH2SO4

0,15molFe2O3-->nH2SO4=0,15.3/1=0,45mol

mH2SO4=nH2SO4.M

               =0,45.(2+32+16.4)

               =44,1g

Giải vậy đc chưa bạn:)

\(a/n_{Fe}=\dfrac{16,8}{56}=0,3mol\\ 4Fe+3O_2\xrightarrow[]{t^0}2Fe_2O_3\\ n_{O_2}=\dfrac{0,3.3}{4}=0,225mol\\ V_{O_2,đktc}=0,225.22,4=5,04l\\ V_{O_2,đkc}=0,225.24,79=5,57775l\\ b/n_{Fe_2O_3}=\dfrac{0,3.2}{4}=0,15mol\\ m_{Fe_2O_3}=0,15.160=24g\)

a) \(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\)

PTHH: \(3Fe+2O_2\xrightarrow[]{t^o}Fe_3O_4\)

            0,3--->0,2----->0,1

\(\Rightarrow m_{Fe_3O_4}=0,1.232=23,2\left(g\right)\)

b) \(V_{O_2}=0,2.22,4=4,48\left(l\right)\Rightarrow V_{kk}=4,48.5=22,4\left(l\right)\)

c) \(n_{O_2\left(hao,h\text{ụt}\right)}=0,2.10\%=0,02\left(mol\right)\)

\(\Rightarrow n_{O_2\left(t\text{ổng}\right)}=0,2+0,02=0,22\left(mol\right)\)

PTHH: \(2KMnO_4\xrightarrow[]{t^o}K_2MnO_4+MnO_2+O_2\)

              0,44<------------------------------------0,22

\(\Rightarrow m_{KMnO_4}=0,44.158=69,52\left(g\right)\)

a.\(n_{KMnO_4}=\dfrac{m_{KMnO_4}}{M_{KMnO_4}}=\dfrac{3,16}{158}=0,02mol\)

\(2KMnO_4\rightarrow\left(t^o\right)K_2MnO_4+MnO_2+O_2\)

0,02                                                       0,01 ( mol )

\(V_{O_2}=n_{O_2}.22,4=0,01.22,4=0,224l\)

b.

\(4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\)

1/75   0,01              1/150   ( mol )

\(m_{Al}=n_{Al}.M_{Al}=\dfrac{1}{75}.27=0,36g\)

\(m_{Al_2O_3}=n_{Al_2O_3}.M_{Al_2O_3}=\dfrac{1}{150}.102=0,68g\)

2KMnO4-to>K2MnO4+MnO2+O2

0,02-------------------------------------0,01

4Al+3O2-to->2Al2O3

\(\dfrac{1}{75}\)---0,01---------\(\dfrac{1}{150}\)

n KMnO4=\(\dfrac{3,16}{158}\)=0,02 mol

=>VO2=0,01.22,4=0,224 l

b)m Al=\(\dfrac{1}{75}\).27=0,36g

=>m Al2O3=\(\dfrac{1}{150}\)102=0,68g

5,6 hay 6,5 :) ?

a, nH2 = 5,6/22,4 = 0,25 (mol)

PTHH: 2H2 + O2 -> (t°) 2H2O

Mol: 0,25 ---> 0,125 ---> 0,25

VO2 = 0,125 . 22,4 = 2,8 (l)

b, mH2O = 0,25 . 18 = 4,5 (g)

c, PTHH: S + O2 -> (t°) SO2

nSO2 = nO2

=> VSO2 = VO2 = 2,8 (l)

a.b.c.

\(n_{H_2}=\dfrac{V}{22,4}=\dfrac{5,6}{22,4}=0,25mol\)

\(2H_2+O_2\rightarrow\left(t^o\right)2H_2O\)

0,25   0,125           0,25  ( mol )

\(V_{O_2}=n.22,4=0,125.22,4=2,8l\)

\(m_{H_2O}=n.M=0,25.18=4,5g\)

d.

\(S+O_2\rightarrow\left(t^o\right)SO_2\)

      0,125         0,125 ( mol )

\(V_{SO_2}=n.22,4=0,125.22,4=2,8l\)