\(n_{O_2}=\dfrac{8}{32}=0.25\left(mol\right)\)

\(CH_4+2O_2\underrightarrow{t^0}CO_2+2H_2O\)

\(0.125....0.25....0.125\)

\(m_{CH_4}=0.125\cdot16=2\left(g\right)\)

\(V_{CO_2}=0.125\cdot22.4=2.8\left(l\right)\)

\(Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3+H_2O\)

\(............0.125.....0.125\)

\(m_{CaCO_3}=0.125\cdot100=12.5\left(g\right)\)

a) C₂H₅OH + 3O₂ --t^o--> 2CO₂ + 3H₂O

b) \(n_{C_2H_5OH}=\dfrac{46}{46}=1\left(mol\right)\)

PTHH: C₂H₅OH + 3O₂ --t^o--> 2CO₂ + 3H₂O

                  1----->3----------->2------->3

=> V_O2 = 22,4.3 = 67,2 (l)

c) m_H2O = 3.18 = 54 (g)

d) V_CO2 = 2.22,4 = 44,8 (l)

a: \(C+O_2\rightarrow CO_2\)(ĐK: t độ)

b: \(n_C=n_{CO_2}=\dfrac{2.4}{12}=0.2\left(mol\right)\)

\(m_{CO_2}=0.2\cdot44=8.8\left(g\right)\)

c: \(n_{O_2}=0.2\left(mol\right)\)

\(\Leftrightarrow V_{O_2}=4.48\left(lít\right)\)

hay \(V_{KK}=22.4\left(lít\right)\)

a. \(n_C=\dfrac{2,4}{12}=0,2\left(mol\right)\)

PTHH : C + O₂ -t^o> CO₂

           0,2     0,2       0,2

b. \(m_{CO_2}=0,2.44=8,8\left(g\right)\)

c. \(V_{O_2}=0,2.22,4=4,48\left(l\right)\)

\(V_{kk}=4,48.5=22,4\left(l\right)\)

n _C=\(\dfrac{2,4}{12}\)=0,2 mol

C+O₂-to>CO₂

0,2--0,2-----0,2 mol

=>m _CO2=0,2.44=8,8g

=>V _kk=0,2.22,4.5=22,4l

\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ pthh:CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)
               0,2    0,4                  0,4 
\(V_{O_2}=0,4.22,4=8,96\left(l\right)\\ m_{H_2O}=0,4.18=7,2\left(g\right)\)

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

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

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

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

Oxi chiếm 20% em nhé !

Oxi chiếm 20%. 

1) \(C+O_2\rightarrow CO_2\\ C+CO_2\rightarrow2CO\) 
2) 
\(pthh:C+O_2\rightarrow CO_2\) 
   => số mol bằng nhau 
\(n_{O_2}=\dfrac{6,4}{16}=0,4\left(mol\right)\)
áp vào pt trên ta có : n_CO2  = 0,4 (mol) 
=> \(m_{CO_2}=0,4.44=17,6\left(g\right)\) 
=> d_CO2/H2 = 44/2 = 22 
d_CO2/H2 = 44/2 = 22

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

n_C = 1,2 / 12 = 0,1 (mol)

n_O2 = 2,24 / 22,4 = 0,1 mol

=> n_CO2 = n_C = n_O2 = 0,1 mol

=> V_CO2(đktc) = 0,1 x 22,4 = 2,24 lít

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

\(2C_2H_2+5O_2\underrightarrow{t^o}4CO_2+2H_2O\)

a, Giả sử: \(\left\{{}\begin{matrix}n_{CH_4}=x\left(mol\right)\\n_{C_2H_2}=y\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow x+y=\dfrac{22,4}{22,4}=1\left(mol\right)\left(1\right)\)

Ta có: \(n_{CO_2}=\dfrac{35,84}{22,4}=1,6\left(mol\right)\)

Theo PT: \(\Sigma n_{CO_2}=n_{CH_4}+2n_{C_2H_2}\)

\(\Rightarrow x+2y=1,6\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,4\left(mol\right)\\y=0,6\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{0,4}{1}.100\%=40\%\\\text{ }\%V_{C_2H_2}=60\%\end{matrix}\right.\)

b, Theo PT: \(\Sigma n_{O_2}=2n_{CH_4}+\dfrac{5}{2}n_{C_2H_2}=2,3\left(mol\right)\)

\(\Rightarrow m_{O_2}=2,3.32=73,6\left(g\right)\)

c, PT: \(CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\)

Theo PT: \(n_{Na_2CO_3}=n_{CO_2}=1,6\left(mol\right)\)

\(\Rightarrow C_{M_{Na_2CO_3}}=\dfrac{1,6}{0,8}=2M\)

Bạn tham khảo nhé!

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