a) nP=0,4(mol)

PTHH: 4P + 5 O2 -to->2 P2O5

Ta có: nP2O5= 2/4. nP= 2/4 . 0,4= 0,2(mol)

=>mP2O5=142.0,2=28,4(g)

b) nO2= 5/4. nP=5/4.0,4=0,5(mol)

=>V(O2,đktc)=0,5.22,4=11,2(l)

Vkk(đktc)=100/20. V(O2,đktc)=100/20 . 11,2= 56(l)

a) nP = 12,4/31=0,4(mol)

PTHH: 4P + 5O2 --> 2P2O5

(mol)  :0,4     0,5          0,2

mP2O5 = 0,2 . 142 = 28,4

vậy sản phẩm thu được là P2O5 28,4 gam

b) VO2 = 0,5 . 22,4 =11,2(l)

Mà VO2 chiếm 20% thể tích không khí

--> Vkk = 100%.11,2/20% = 1120%/20% = 56(l)

Vậy Vkk(đktc) = 56(l) 

\(n_{Zn}=\dfrac{130}{65}=2mol\\ Zn+2HCl\rightarrow ZnCl_2+H_2\\ n_{Zn}=n_{H_2}=2mol\\ V_{H_2}=2.22,4=44,8l\\ 1000ml=1l\\ n_{HCl}=2.2=4mol\\ C_{M_{HCl}}=\dfrac{4}{1}=4M\)

\(n_{Zn}=\dfrac{m_{Zn}}{M_{Zn}}=\dfrac{130}{65}=2mol\) 

PTHH: Zn + 2HCl \(\rightarrow\) ZnCl₂ + H₂

TL;       1        2          1          1

mol:      2 \(\rightarrow\) 4

\(m_{HCl}=n.M=4.36,5=146g\) 

đổi 1000 ml= 1l

\(C\%_{ddHCl}=\dfrac{m_{HCl}}{V_{HCl}}.100\%=\dfrac{146}{1}.100=14600\%\)

số hơi lớn em xem lại đề nhé

\(n_{CO_2}=\dfrac{5,6}{22,4}=0,25mol\)

\(\left\{{}\begin{matrix}CH_4:x\left(mol\right)\\C_2H_4:y\left(mol\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}16x+28y=3,6\\BTC:x+2y=n_{CO_2}=0,25\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=0,05mol\\y=0,1mol\end{matrix}\right.\)

a)\(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)

\(C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\)

b)\(m_{CH_4}=0,05\cdot16=0,8g\)

\(m_{C_2H_4}=0,1\cdot28=2,8g\)

c)\(\Sigma n_{O_2}=2n_{CH_4}+3n_{C_2H_4}=2\cdot0,05+3\cdot0,1=0,4mol\)

\(\Rightarrow V_{O_2}=0,4\cdot22,4=8,96l\)

\(\Rightarrow V_{kk}=5V_{O_2}=5\cdot8,96=44,8l\)

C+O2​to-->​CO2​

0,3--0,3--0,3

n_C = 3,6 / 12 = 0,3 (mol)

=> V_CO2(đktc) = 0,3 x 22,4 =6,72lít

=>Vkk=6,72\5=33,6l

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

a) Ta có: \(n_C=\dfrac{3,6}{12}=0,3\left(mol\right)=n_{CO_2}\) 

\(\Rightarrow V_{CO_2}=0,3\cdot22,4=6,72\left(l\right)\)

b) Theo PTHH: \(n_{O_2}=n_C=0,3mol\) 

\(\Rightarrow V_{O_2}=6,72\left(l\right)\) \(\Rightarrow V_{kk}=6,72\cdot5=33,6\left(l\right)\)

TK

Từ C₂H₄O₂ ta có: M = 60 g/mol; m_C = 2 x 12 = 24 g; m_H = 4 x 1 = 4 g;

M_O = 2 x 16 = 32 g.

%C = (24 : 60) x 100% = 40%; %H = (4 : 60) x 100% = 6,67%;

%O = 100% - 40% - 6,67% = 53,33%.

a.b.\(n_{Fe_2O_3}=\dfrac{8}{160}=0,05\left(mol\right)\)

\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

\(Fe_2O_3+3H_2\rightarrow\left(t^o\right)2Fe+3H_2O\)

0,05  >    0,1                                  ( mol )

\(\dfrac{1}{30}\)          0,1               \(\dfrac{1}{15}\)                  ( mol )

\(m_{thu.được}=m_{Fe_2O_3\left(dư\right)}+m_{Fe}\)

               \(=\left(0,05-\dfrac{1}{30}\right).160+\dfrac{1}{15}.56=6,4\left(g\right)\)

c. Để khử hết oxit sắt thì thể tích H₂ cần dùng

\(Fe_2O_3+3H_2\rightarrow\left(t^o\right)2Fe+3H_2O\)

 0,05         0,15                          ( mol )

\(V_{H_2}=0,15.22,4=3,36\left(l\right)\)

a) \(n_{HCl}=\dfrac{1,095}{36,5}=0,03\left(mol\right)\)

b) \(n_{CH_4}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

`C_2H_4+Br_2->C_2H_4Br_2`

`0,09---0,09mol`

`n(Br_2)=(14,4)/160=0,09mol`

`->%(C_2H_4)=((0,09.22,4)/(3,36)).100%=60%`

`->%(CH_4)=100-60=40%`

#yBTrCute