\(n_{O_2}=\dfrac{43.2}{32}=1.35\left(mol\right)\)

\(2KClO_3\underrightarrow{^{t^0}}2KCl+3O_2\)

\(0.9...........................1.35\)

\(H\%=\dfrac{0.9}{1}\cdot100\%=90\%\)

PT: \(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)

______1_____________1,5 (mol)

⇒ m_{O2 (lí thuyết)} = 1,5.32 = 48 (g)

Mà: m_{O2 (thực tế)} = 43,2 (g)

\(\Rightarrow H\%=\dfrac{43,2}{48}.100\%=90\%\)

Bạn tham khảo nhé!

\(n_{O_2}\)=\(\dfrac{43,2}{32}=1,35\left(mol\right)\)

PTHH 2KClO₃-----t^o--->2KCl +3O₂

=>\(n_{O_2\left(lt\right)}=1.\dfrac{3}{2}=1,5\left(mol\right)\)

=>H%=\(\dfrac{1,35}{1,5}.100\%=90\%\)

2KClO3--->2KCl+302

nKClO3=122,5:122,5=1 mol

nO2=43,2:32=1,35 mol

So sánh nKClO3/2     >  nO2      --->nO2 hết KClO3 dư  tính theo O2

                 (=0,5)             (=0,45)

Theo pt -->nKClO3 thực =nO2=1,35 

mKClO3=1,35.122,5=165,375 g

H=(165,375/122.5).100%=135%

Mình nhầm bạn nhá phải là nKClO3=nO2.2/3=1,35.2/3=0,9 mol

g KClO3 tt=0,9 mol

H=(0,9/1).100%=90%

\(n_{O_2}=\frac{43,2}{32}=1,35mol\)

PTHH: \(2KClO_3\rightarrow^{t^o}2KCl+3O_2\uparrow\)

Theo phương trình \(n_{KClO_3\left(p/ứ\right)}=\frac{2}{3}n_{O_2}=0,9mol\)

\(\rightarrow H=\frac{0,9.122,5}{122,5}.100\%=90\%\)

\(n_{Cu}=\dfrac{6.4}{64}=0.1\left(mol\right)\)

\(2Cu+O_2\underrightarrow{^{t^0}}2CuO\)

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

\(m_{CuO\left(tt\right)}=0.1\cdot80=8\left(g\right)\)

\(H\%=\dfrac{m_{lt}}{m_{tt}}\cdot100\%=\dfrac{6.4}{8}\cdot100\%=80\%\)

nAl=10/27(mol)

ta ccó pthh: 2Al+3S->Al2S3( nhiệt dộ cao)

theo ptth=> nAl2S3(lý thuyết)=1/2.nAl=\(\dfrac{1}{2}.\dfrac{10}{27}\)=\(\dfrac{5}{27}\)(mol)

=> mAl2S3(lý thuyết)=\(\dfrac{5}{27}.150=\dfrac{250}{9}\)(g)

=>H=\(\dfrac{mAL2S3\left(thucte\right)}{mAL2S3\left(lythuyet\right)}.100\%=\dfrac{25,5}{\dfrac{250}{9}}=91,8\%\)

\(n_{Al}=\dfrac{10}{27}\left(mol\right)\)

\(2Al+3S\underrightarrow{^{t^0}}Al_2S_3\)

\(\dfrac{10}{27}...........\dfrac{5}{27}\)

\(m_{Al_2S_3\left(lt\right)}=\dfrac{5}{27}\cdot150=\dfrac{250}{9}\left(g\right)\)

\(H\%=\dfrac{m_{tt}}{m_{lt}}\cdot100\%=\dfrac{25.5}{\dfrac{250}{9}}\cdot100\%=91.8\%\)

Gọi hiệu suất là

Ta có :

n CuO = 20/80 = 0,25(mol)

n Cu = n CuO(tt) = 0,25a(mol)

=> n CuO dư = 0,25 - 0,25a(mol)

$CuO + H_2 \xrightarrow{t^o} Cu + H_2O$

Suy ra :

0,25a.64 + (0,25 - 0,25a).80 = 16,8

=>a = 0,8 = 80%

nKClO3=0,1(mol)

PTHH: 2 KClO3 -to-> 2 KCl +3 O2

0,1_____________0,1______0,15(mol)

a) mKCl=0,1.74,5=7,45(g)

b) V(O2,đktc)=0,15.22,4=3,36(l)

\(n_{KClO_3}=\dfrac{12,25}{122,5}=0,1\left(mol\right)\)

PTHH: \(2KClO_3\rightarrow2KCl+3O_2\)

               0,1             0,1         0,15    (mol)

\(m_{KCl}=0,1.74,5=7,45\left(g\right)\)

\(V_{O_2}=0,15.22,4=3,36\left(l\right)\)

PTHH: \(CaCO_3\rightarrow Cao+CO_2\)

\(1mol\rightarrow1mol\)

\(100g\rightarrow65g\)

Theo bài: 4 tấn \(\rightarrow\)m tấn

Khối lượng \(CaO\)được theo lý thuyết là:

\(m_{CaOlt}\)=\(\dfrac{4.56}{100}=2,24\)(tấn)

Hiệu suất phản ứng là:

\(H=\dfrac{1,68}{22,4.}100\%=75\%\)

a.\(n_{KClO_3}=\dfrac{m_{KClO_3}}{M_{KClO_3}}=\dfrac{12,25}{122,5}=0,1mol\)

\(2KClO_3\rightarrow\left(t^o\right)2KCl+3O_2\)

    2                       2           3       ( mol )

  0,1                                 0,15

\(V_{O_2}=n_{O_2}.22,4=0,15.22,4=3,36l\)

b.\(V_{kk}=V_{O_2}.5=3,36.5=16,8l\)

c.\(n_{Fe}=\dfrac{m_{Fe}}{M_{Fe}}=\dfrac{28}{56}=0,5mol\)

\(3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\)

  3        2                   1         ( mol )

0,5   >   0,15                          ( mol )

0,225              0,15                 ( mol )

\(m_{Fe\left(du\right)}=n_{Fe\left(du\right)}.M_{Fe}=\left(0,5-0,225\right).56=15,4g\)

Cập nhật lại câu hỏi đi bạn :v