\(n_{Cl_2}=\dfrac{11.2}{22.4}=0.5\left(mol\right)\)

\(2KMnO_4+16HCl_{\left(đ\right)}\rightarrow2KCl+2MnCl_2+5Cl_2+8H_2O\)

\(0.2...................1.6...........................................0.5\)

\(m_{KMnO_4}=0.2\cdot158=31.6\left(g\right)\)

\(V_{dd_{HCl}}=\dfrac{1.6}{1}=1.6\left(l\right)=1600\left(ml\right)\)

\(n_{Cl_2}=\dfrac{11,2}{22,4}=0,5(mol)\\ PTHH:2KMnO_4+16HCl\to 2KCl+2MnCl_2+5Cl_2+8H_2O\\ \Rightarrow n_{KMnO_4}=\dfrac{2}{5}n_{Cl_2}=0,2(mol);n_{HCl}=\dfrac{16}{5}n_{Cl_2}=1,6(mol)\\ \Rightarrow m_{KMnO_4}=0,2.158=31,6(g);V_{dd_{HCl}}=\dfrac{1,6}{1}=1,6(l)=1600(ml)\)

2Al+6HCl->2AlCl3+3H2

0,5----1,5-------------------------0,75

n H2=0,075 mol

cần 10 quả bóng =>0,075.10=0,75 mol

=>mAl=0,5.27=13,5g

m HCl=1,5.36,5=54,75g

=>m dd=75g

\(n_{H_2}=\dfrac{1,68}{22,4}=0,075\left(mol\right)\\ pthh:2Al+6HCl\rightarrow2AlCl_3+3H_2\) 
           0,1125   0,15                     0,075 
\(m_{Al}=0,1125.27=3,0375\left(g\right)\\ m_{HCl}=\dfrac{\left(0,15.36,5\right).100}{7,3}=75\left(g\right)\) 

a)

SO3 + H2O $\to$ H2SO4

n H2SO4 = n SO3 = 240/80 = 3(mol)

m H2SO4 = 3.98 = 294(gam)

b)

2H2 + O2 $\xrightarrow{t^o}$ 2H2O

V H2 / 2 = 0,56 < V O2 = 1,68 nên O2 dư

n H2O = n H2 = 1,12/22,4 = 0,05(mol)

m H2O = 0,05.18 = 0,9(gam)

V H2O = m/D = 0,9/1 = 0,9(ml)

c)

n NaOH = 16/40 = 0,4(mol)

Na2O + H2O $\to$ 2NaOH

n Na2O = 1/2 n NaOH = 0,2(mol)
m Na2O = 0,2.62 = 12,4(gam)

a)Ta có PT:SO4+H2O=H2SO4

Theo bài ra t có n_{SO4=240:96=2,5}

Mà n_So4=n_H2So4=2,5 mol

suy ra:m_H2So4=2,5.98=245g

ta có PT 2H2+O2=2H2O

n_H2=1,12:22,4=0,05MOL

n_O2=1,68:22.4=0,075mol

theo pt ta có 0,05:2<0,075:1

suy ra H2 hết,O2 dư.

theo PT ta có:n_H2O=nH2=0,05mol

suy ra V_H2O=.....

m_H2O=.....

a) PTHH: NaOH + Al + H2O -> NaAlO2 + 3/2 H2

b) nH2= 0,6(mol)

-> nAl=0,4(mol) => mAl=0,4.27=10,8(g)

c) nAl=0,18((mol); nNaOH=0,2(mol)

PTHH: 0,18/1 < 0,2/1

=> Al hết, NaOH dư, tính theo nAl.

-> nH2= 3/2. 0,18=0,27(mol)

=>V(H2,đktc)=0,27.22,4= 6,048(l)

\(n_{H_2}=\dfrac{13.44}{22.4}=0.6\left(mol\right)\)

\(2NaOH+2Al+2H_2O\rightarrow2NaAlO_2+3H_2\)

\(...........0.4.........................0.6\)

\(m_{Al}=0.4\cdot27=10.8\left(g\right)\)

\(n_{Al}=\dfrac{4.86}{27}=0.18\left(mol\right)\)

\(n_{NaOH}=\dfrac{8}{40}=0.2\left(mol\right)\)

\(2NaOH+2Al+2H_2O\rightarrow2NaAlO_2+3H_2\)

\(2.................2\)

\(0.2...............0.18\)

\(LTL:\dfrac{0.2}{2}>\dfrac{0.18}{2}\)

\(\Rightarrow NaOHdư\)

\(n_{H_2}=0.18\cdot\dfrac{3}{2}=0.27\left(mol\right)\)

\(V_{H_2}=0.27\cdot22.4=6.048\left(l\right)\)

a, PTHH: 2NaOH + 2Al + 2H₂O → 2NaAlO₂ + 3H₂↑

b, nH2=0,6 mol -mAl=10,8 g 

c, nAl=0,16(592) mol 

   nNaOH=0,2 mol 

-NaOH hết, Al dư 

-VH2=6,72 l

\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ n_{H_2SO_4}=n_{H_2}=\dfrac{1,568}{22,4}=0,07\left(mol\right)\\ n_{H_2SO_4\left(ban.đầu\right)}=5.0,3=1,5\left(mol\right)\\ V_{ddH_2SO_4\left(lấy\right)}=\dfrac{0,07}{1,5}.3=0,14\left(l\right)=140\left(ml\right)\)

\(n_{FeCl_3}=\dfrac{16.25}{162.5}=0.1\left(mol\right)\)

\(Fe+\dfrac{3}{2}Cl_2\underrightarrow{t^0}FeCl_3\)

\(......0.15......0.1\)

\(2KMnO_4+16HCl\rightarrow2KCl+2MnCl_2+5Cl_2+8H_2O\)

\(0.06...............0.48........................................0.15\)

\(m_{KMnO_4}=0.06\cdot158=9.48\left(g\right)\)

\(V_{dd_{HCl}}=\dfrac{0.48}{1}=0.48\left(l\right)=480\left(ml\right)\)

\(2Fe+ 3Cl_2 \xrightarrow{t^o} 2FeCl_3\\ n_{Cl_2} = \dfrac{3}{2}n_{FeCl_3} = \dfrac{3}{2}.\dfrac{16,25}{162,5} = 0,15(mol)\\ 2KMnO_4 + 16HCl \to 2KCl + 2MnCl_2 + 5Cl_2 + 8H_2O\\ n_{KMnO_4} = \dfrac{2}{5}n_{Cl_2} = 0,06(mol)\\ \Rightarrow m_{KMnO_4} = 0,06.158 = 9,48(gam)\\ n_{HCl} = \dfrac{16}{4}n_{Cl_2} = 0,48(mol)\\ \Rightarrow V_{dd\ HCl} = \dfrac{0,48}{1} = 0,48(lít) = 480(ml)\)

Phương trình hóa học của phản ứng:

3Cl2 + 2Fe → 2FeCl3

Theo pt: 

2KMnO4 + 16HCl → 2KCl + 2MnCl2 + 5Cl2 + 8H2O

Theo pt:

mKMnO4 cần = 0,06. 158 = 9,48g