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

\(n_C=n_{CO_2}=\dfrac{22}{44}=0,5mol\)

\(\Rightarrow m_C=0,5.12=6\left(g\right)\)

\(n_{CO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ n_{Fe_3O_4}=\dfrac{23,2}{232}=0,1\left(mol\right)\\ 4CO+Fe_3O_4\rightarrow\left(t^o\right)3Fe+4CO_2\\ V\text{ì}:\dfrac{0,1}{1}>\dfrac{0,2}{4}\Rightarrow Fe_3O_4d\text{ư}\\ \Rightarrow n_{Fe}=\dfrac{3}{4}.0,2=0,15\left(mol\right)\\ n_{Fe_3O_4\left(d\text{ư}\right)}=0,1-\dfrac{0,2}{4}=0,05\left(mol\right)\\ m_{r\text{ắn}}=m_{Fe_3O_4\left(d\text{ư}\right)}+m_{Fe}=0,05.232+0,15.56=20\left(g\right)\)

a) Fe₂O₃ + 3CO --t^o--> 3CO₂ + 2Fe

b) Theo ĐLBTKL: m_Fe2O3 + m_CO = m_CO2 + m_Fe

=> m_CO2 = 16+8,4 - 11,2 = 13,2(g)

c)
\(n_{CO_2}=\dfrac{13,2}{44}=0,3\left(mol\right)\)

V_CO2 = 0,3.22,4 = 6,72 (l)

Số phân tử CO2 = 0,3.6.10²³ = 1,8.10²³

\(a,Fe_2O_3+3CO\xrightarrow{t^o}2Fe+3CO_2\\ b,BTKL:m_{Fe_2O_3}+m_{CO}=m_{Fe}+m_{CO_2}\\ \Rightarrow m_{CO_2}=16+8,4-11,2=13,2(g)\\ c,n_{CO_2}=\dfrac{13,2}{44}=0,3(mol)\\ V_{CO_2}=0,3.22,4=6,72(l)\\ \text{Số phân tử }CO_2:0,3.6.10^{23}=1,8.6.10^{23}\)

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

C+O2-to>CO2

0,6-------------0,6

CO2+C-to>2CO

n C=0,7 mol

n O2=0,6 mol

C dư

=>m CO2=0,6.44=26,4g

=>dhh\H2=\(\dfrac{44}{2}=22\)

a) PTHH:FeO + CO -to-> Fe + CO2

b) Theo ĐLBTKL, ta có:

mFeO+ mCO= mFe + mCO2

=> mCO2=(mFeO + mCO)-mFe= (14+36)-28=22(g)

C+O2→CO₂

+n_C=\(\dfrac{4,8}{12}=0,4\left(mol\right)\)

+Theo PTHH ta có: 

-n_CO2=n_C=0,4(mol)

+m_CO2=0,4.44=17,6(gam)

a)

\(C + O_2 \xrightarrow{t^o} CO_2\\ n_{CO_2} = n_{O_2} = \dfrac{6,4}{32} = 0,2(mol)\\ \Rightarrow m_{CO_2} = 0,2.44 = 8,8(gam)\)

b)

\(n_C = \dfrac{6}{12} = 0,5(mol)\\ n_{O_2} =\dfrac{19,2}{32} = 0,6(mol)\\ C + O_2 \xrightarrow{t^o} CO_2\)

\(n_C = 0,5 < n_{O_2} = 0,6 \Rightarrow\) Oxi dư.

\(n_{CO_2} = n_C = 0,5(mol)\\ \Rightarrow m_{CO_2} = 0,5.44 = 22(gam)\)

PTHH:

\(C_2+2O_2\underrightarrow{t^o}2CO_2\)

a/ Ta có: \(n_{O_2}=\dfrac{6.4}{32}=0.2\left(mol\right)\)

\(C_2+2O_2\underrightarrow{t^o}2CO_2\)

         2         2

         0.2       x

\(=>x=\dfrac{0.2\cdot2}{2}=0.2=n_{CO_2}\)

\(=>m_{CO_2}=0.2\cdot8.8\left(g\right)\)

 C+ O2  -to--> CO2

a)   nO2= m/n= 6,4/ 32= 0.2 mol

     0.2 mol O2 -------> 0.2 mol CO2

     m CO2= m x n= 0.2 x  44 = 8.8 gam

BTKL
m C + m O2 = m CO2

=> 24 + m O2 = 88

=> m O2 = 64 ( g )