m than đá= 9.80%=7,2(kg)
7,3kg = 7200(g)
n C = 7200/12=600(mol)
C+O2---->CO2
600---------600(mol)
V CO2=600.22,4=13440(l)
Chúc bạn hcoj tôt
\(m_{than}=9.\left(100\%-20\%\right)=7,2\left(kg\right)\)
\(n_C=\frac{7,2}{12}=0,6\left(kmol\right)\)
\(C+O2\underrightarrow{t^o}CO2\)
0,6_________0,6(kmol)
\(V_{CO_2}=0,6.1000.22,4=13440\left(l\right)\)
m than đá=9.(100%−20%)=7,2(kg)
nC=7,2\12=0,6(kg\mol)
C+O2to→CO2
0,6_________0,6(kg\mol)
=>VCO2=0,6.1000.22,4=13440(l)