\(n_C=0,5\left(mol\right)\\ C+O_2\rightarrow\left(t^o\right)CO_2\\ n_{O_2}=n_C=0,5\left(mol\right)\\ \Rightarrow V_{O_2\left(đktc\right)}=0,5.22,4=11,2\left(l\right)\\ Mà:V_{O_2}=\dfrac{1}{5}V_{kk}\Leftrightarrow V_{kk}=5.V_{O_2}=5.11,2=56\left(lít\right)\)

nC = 2.4/12 = 0.2 mol 

C + O2 -to-> CO2 

0.2__0.2 

VKK = 5VO2 = 5*0.2*22.4 = 22.4 (l) 

Câu 1:

\(n_{CO_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

PTHH: C + O₂ --t^o--> CO₂

         0,15<-----------0,15

=> \(\%C=\dfrac{0,15.12}{2}.100\%=90\%\)

Câu 2:

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

             0,2<-------------------0,3

=> m_Al = 0,2.27 = 5,4 (g)

Câu 1.

\(n_{CO_2}=\dfrac{3,36}{22,4}=0,15mol\)

\(C+O_2\underrightarrow{t^o}CO_2\)

0,15         0,15

\(m_C=0,15\cdot12=1,8g\)

\(\%C=\dfrac{1,8}{2}\cdot100\%=90\%\)

Câu 2.

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3mol\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)

 0,2                                    0,3

\(m_{Al}=0,2\cdot27=5,4g\)

\(n_C=\dfrac{24}{12}=2\left(mol\right)\)

\(C+O_2\underrightarrow{^{t^0}}CO_2\)

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

\(V_{kk}=5V_{O_2}=5\cdot2\cdot22.4=224\left(l\right)\)

Oxi chiếm 20% em nhé !

Oxi chiếm 20%. 

C+O₂-to>CO₂

1,5--1,5 -----1,5mol

n C=\(\dfrac{18}{12}\)=1,5 mol

=>V_kk=1,5.22,4.5=168l

=>V_CO2=1,5.22,4=33,6l

nC = 18/12 = 1,5 (mol)

PTHH: C + O2 -> (t°) CO2

Mol: 1,5 ---> 1,5 ---> 1,5 

VO2 = 1,5 . 22,4 = 33,6 (l)

Vkk = 33,6 . 5 = 168 (l)

VCO2 = 1,5 . 22,4 = 33,6 (l)

\(n_S=\dfrac{3.2}{32}=0.1\left(mol\right)\)

\(n_{O_2}=\dfrac{1.12}{22.4}=0.05\left(mol\right)\)

\(S+O_2\underrightarrow{t^0}SO_2\)

\(0.05.0.05...0.05\)

\(\Rightarrow Sdư\)

\(V_{SO_2}=0.05\cdot22.4=1.12\left(l\right)\)

\(b.\)

\(S+O_2\underrightarrow{t^0}SO_2\)

\(0.1..0.1\)

\(V_{kk}=5V_{O_2}=5\cdot0.1\cdot22.4=11.2\left(l\right)\)

Tks bạn

a, PT: \(S+O_2\underrightarrow{t^o}SO_2\)

Ta có: \(n_S=\dfrac{3,2}{32}=0,1\left(mol\right)\)

\(n_{O_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,1}{1}>\dfrac{0,05}{1}\), ta được S dư.

Theo PT: \(n_{SO_2}=n_{O_2}=0,05\left(mol\right)\) \(\Rightarrow V_{SO_2}=0,05.22,4=1,12\left(l\right)\)

b, Theo PT: \(n_{O_2}=n_S=0,1\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,1.22,4=2,24\left(l\right)\)

\(\Rightarrow V_{kk}=2,24.5=11,2\left(l\right)\)

a, Theo giả thiết ta có: \(n_P=\dfrac{3,1}{31}=0,1\left(mol\right)\)

\(4P+5O_2--t^o->2P_2O_5\)

Ta có: \(n_{O_2}=\dfrac{5}{4}.n_P=0,125\left(mol\right)\Rightarrow V_{O_2\left(đktc\right)}=0,125.22,4=2,8\left(l\right)\)

b, Theo giả thiết ta có: \(n_{CH_4}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

\(CH_4+2O_2--t^o->CO_2+2H_2O\)

Ta có: \(n_{O_2}=2.n_{CH_4}=0,1\left(mol\right)\Rightarrow V_{O_2\left(đktc\right)}=2,24\left(l\right)\)