nCO2 = 8,96/22,4 = 0,4 (mol)

PTHH: C + O2 -> (t°) CO2

Mol: 0,4 <--- 0,4 <--- 0,4

mO2 = 0,4 . 32 = 12,8 (g)

mC = 0,4 . 12 = 4,8 (g)

%mC = 4,8/80 = 24%

%tạp chất = 100% - 24% = 76%

.

\(V_{O_2}=9,52m^3=9520l\)

\(n_{O_2}=\dfrac{V_{O_2}}{22,4}=\dfrac{9520}{22,4}=425mol\)

\(C+O_2\rightarrow\left(t^o\right)CO_2\)

425 425           425 ( mol )

\(m_{CO_2}=n_{CO_2}.M_{CO_2}=425.44=18700g\)

\(m_C=n_C.M_C=425.12=5100g\)

\(m_{than}=\dfrac{5100.100}{\left(100-25\right)}=6800g\)

Đổi 9,52m3 = 9520 lít

nO2 = 9520/22,4 = 425 (mol)

PTHH: C + O2 -> (t°) CO2

Mol: 425 <--- 425 ---> 425

mCO2 = 425 . 44 = 18700 (g)

mC = 425 . 12 = 5100 (g)

m = 5100 : (100% - 25%) = 6800 (g)

Đổi 2,5kg = 2500g

mC = 2500 . (100% - 16%) = 2100 (g)

nC = 2100/12 = 175 (mol)

PTHH: C + O2 -> (t°) CO2

Mol: 175 ---> 175 ---> 175

VO2 = 175 . 22,4 = 3920 (l)

mCO2 = 44 . 175 = 7700 (g)

\(m_C=\dfrac{2,5.\left(100-16\right)}{100}=2,1kg\)

\(m_C=2,1kg=2100g\)

\(n_C=\dfrac{m_C}{M_C}=\dfrac{2100}{12}=175mol\)

\(C+O_2\rightarrow\left(t^o\right)CO_2\)

175  175           175    ( mol )

\(V_{O_2}=n_{O_2}.22,4=175.22,4=3920l\)

\(m_{CO_2}=n_{CO_2}.M_{CO_2}=175.44=7700g\)

Phần trăm khối lượng cacbon có trong than đá:

\(100\%-16\%=84\%\)

Khối lượng cacbon:

\(m_C=\dfrac{2,5\cdot84\%}{100\%}=2,1kg\Rightarrow n_C=175mol\)

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

175  175   175

\(V_{O_2}=175\cdot22,4=3920ml=3,92l\)

\(m_{CO_2}=175\cdot44=7700g=7,7kg\)

Bạn chia nhỏ câu hỏi ra được không:v, nhiều thế này mn ngại làm

a. \(n_{CH_4}=\dfrac{10.08}{22,4}=0,45\left(mol\right)\)

PTHH : CH₄ + 2O₂ ----t^o---> CO₂ + 2H₂O

         0,45    0,9              0,45

\(V_{O_2}=0,9.22,4=20,16\left(l\right)\)

\(V_{kk}=20,16.5=100,8\left(l\right)\)

b. \(m_{CO_2}=0,45.44=19,8\left(g\right)\)

c. PTHH : 2KMnO₄ -> K₂MnO₄ + MnO₂ + O₂

              1,8                               0,9

\(m_{KMnO_4}=1,8.158=284,4\left(g\right)\)

Bài 2:

Ta có n_Fe3O4 = 

Bài 1:

Ta có n_CH4 = 

a. \(n_{Al}=\dfrac{27}{27}=1\left(mol\right)\)

\(n_{Al_2O_3}=\dfrac{40.8}{102}=0,4\left(mol\right)\)

PTHH : 4Al + 3O₂ -----t^o---> 2Al₂O₃

              0,8     0,6                   0,4

Xét tỉ lệ : 1 > 0,4 => Al dư , Al₂O₃ đủ

\(m_{Al}=0,8.27=21,6\left(g\right)\)

b. \(m_{tạp.chất}=27-21,6=5,4\left(g\right)\)

\(\%_{tạp.chất}=\dfrac{5.4}{27}.100\%=20\%\)

\(V_{O_2}=20\%V_{kk}=20\%\cdot67,2=13,44l\)

\(\Rightarrow n_{O_2}=\dfrac{13,44}{22,4}=0,6mol\)

\(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)

0,9      0,6     0,3

\(m_{Fe}=0,9\cdot56=50,4g\)

\(m_{Fe_3O_4}=0,3\cdot232=69,6g\)

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

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

Ta có: \(n_{CH_4}+n_{C_2H_4}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\left(1\right)\)

Theo PT: \(n_{CO_2}=n_{CH_4}+2n_{C_2H_4}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{CH_4}=0,2\left(mol\right)\\n_{C_2H_4}=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{0,2.22,4}{6,72}.100\%\approx66,67\%\\\%V_{C_2H_4}\approx33,33\%\end{matrix}\right.\)

b, Theo PT: \(n_{O_2}=2n_{CH_4}+3n_{C_2H_4}=0,7\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,7.22,4=15,68\left(l\right)\)

nP = 15,5 : 31= 0,5 (mol ) 
nO2 = (56 : 22,4 ) . 21% = 0,525 (mol ) 
pthh : 4P+5O2-t--> 2P2O5 
LTL 
0,5 / 4   > 0,525/5 => P dư 
theo pthh , nP2O5 = nO2 = 0,525 (mol) 
=> mP2O5 = 0,525 . 142 = 74,55 (g)

\(n_P=\dfrac{m}{M}=\dfrac{15,5}{31}=0,5\left(mol\right)\)

\(n_{O_2}=\dfrac{V}{22,4}=\dfrac{56\times20}{22,4\times100}=0,05\left(mol\right)\)

\(PTHH:4P+5O_2\rightarrow2P_2O_5\)

Ta có: \(\dfrac{0,5}{4}=0,125>\dfrac{0,05}{5}=0,01\)

→ Sau pư O₂ hết, P dư

→ Theo \(n_{O_2}\)

Theo pthh: \(n_{P_2O_5}=\dfrac{2}{5}n_{O_2}=\dfrac{2}{5}.0,05=0,02\left(mol\right)\)

\(m_{P_2O_5}=n.M=0,02\times142\times80\%=1,36\left(g\right)\)

\(n_{P\left(phảnứng\right)}=\dfrac{4}{5}n_{O_2}=\dfrac{4}{5}.0,05=0,04\left(mol\right)\)

\(m_{P\left(phảnứng\right)}=n.M=0,04.31=1,24\left(g\right)\)

\(m_{P\left(dư\right)}=15,5-1,24=14,26\left(g\right)\)