a.\(C\%_{KCl}=\dfrac{20}{600}.100=3,33\%\)

b.2,5kg = 2500g

\(C\%_{Al_2\left(SO_4\right)_3}=\dfrac{34,2}{2500}.100=1,368\%\)

`(a): C%_[KCl] = 20 / 600 . 100 ~~ 3,33 %`

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`(b):` Đổi `2,5 kg = 2500 g`

`C%_[Al_2 (SO_4)_3] = [ 34,2 ] / 2500 . 100 = 1,368%`

\(n_{H_2}=\dfrac{0,224}{22,4}=0,01\left(mol\right)\\ pthh:2R+2H_2O\rightarrow2ROH+H_2\) 
          0,02                                0,01 (mol) 
\(\Rightarrow M_R=0,78:0,02=39\left(\dfrac{g}{mol}\right)\) 
mà R hóa trị I => R là K

1)

$a)m_{Cl_2} = 0,3.71 = 21,3(gam)$
$b)V_{Cl_2} = 0,3.22,4 = 6,72(lít)$
Số phân tử $= 0,3.6.10^{23} = 1,8.10^{23}$ phân tử

2)

$a)m_{CO_2} = 0,45.44 = 19,8(gam)$
$b) V_{CO_2} = 0,45.22,4 = 10,08(lít)$
Số phân tử $= 0,45.6.10^{23} = 2,7.10^{23}$ phân tử

\(1.a.2Mg+O_2-^{t^o}\rightarrow2MgO\\ b.Fe+2AgNO_3\rightarrow Fe\left(NO_3\right)_2+2Ag\\ c.C_2H_4+3O_2-^{t^o}\rightarrow2CO_2+2H_2O\\ d.CuO+2HCl\rightarrow CuCl_2+H_2O\\ e.2Na+2H_2O\rightarrow2NaOH+H_2\\ f.4Al+3O_2-^{t^o}\rightarrow2Al_2O_3\)

\(2.a.Magie+Axitclohidric\rightarrow MagieClorua+Hidro\\ b.Mg+2HCl\rightarrow MgCl_2+H_2\\ c.m_{Mg}+m_{HCl}=m_{MgCl_2}+m_{H_2}\\ d.m_{HCl}=m_{MgCl_2}+m_{H_2}-m_{Mg}=47,5+1-12=36,5\left(g\right)\)

\(4Na+O_2\rightarrow2Na_2O\)

\(Ca+\dfrac{1}{2}O_2\rightarrow CaO\)

\(S+O_2\xrightarrow[]{t^o}SO_2\)

\(Na_2O+H_2O\rightarrow2NaOH\)

\(K+H_2O\rightarrow KOH+\dfrac{1}{2}H_2\uparrow\)

\(SO_3+H_2O\rightarrow H_2SO_4\)

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

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

\(H_2+\dfrac{1}{2}O_2\xrightarrow[]{t^o}H_2O\)

\(CuO+H_2\xrightarrow[]{t^o}Cu+H_2O\)

\(H_2+PbO\xrightarrow[]{t^o}Pb+H_2O\)

n H2=\(\dfrac{11,2}{22,4}\)=0,5 mol

=>m H2=0,5.2=1 g

ĐLBTKLg:

=>a=m Fe=30,2+1-9=22,2g

nO2 = 5,6/22,4 = 0,25 (mol)

PTHH: 2KClO3 -> (t°, MnO2) 2KCl + 3O2

nKClO3 = 0,25 : 3 . 2 = 1/6 (mol)

nKClO3 = 1/6 . 122,5 = 245/12 (g)

nMg = 2,4/24 = 0,1 (mol)

PTHH: 2Mg + O2 -> (t°) 2MgO

LTL: 0,1/2 < 0,25 => O2 dư

nMgO = 0,1 (mol)

nMgO = 0,1 . 40 = 4 (g)

nO2 = 5,6 : 22,4 = 0,25(mol) 
pthh : 2KClO3 -t--> 2KCl + 3O2
           1/6     <----------------0,25(mol)
=>mKClO3 = 1/6.114,5=229/12(g)

nMg=2,4:24=0,1(mol)
pthh 2Mg+O2 -t-> 2MgO
               0,1--------->0,1(mol) 
=> mMgO = 0,1.40=4 (g)