PTHH: \(C+O_2\xrightarrow[]{t^o}CO_2\)
Ta có: \(\left\{{}\begin{matrix}n_C=\dfrac{6,4}{12}=\dfrac{8}{15}\left(mol\right)\\n_{O_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) C còn dư
\(\Rightarrow\left\{{}\begin{matrix}n_{C\left(dư\right)}=\dfrac{7}{30}\left(mol\right)\\n_{CO_2}=0,3\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{C\left(dư\right)}=\dfrac{7}{30}\cdot12=2,8\left(g\right)\\V_{CO_2}=0,3\cdot22,4=6,72\left(l\right)\end{matrix}\right.\)
nCnC == 4,812=0,4(mol)4,812=0,4(mol)
nO2nO2 == 6,7222,4=0,3(mol)6,7222,4=0,3(mol)
PTHH: C+O2C+O2 to→→to CO2CO2
Do: 0,4>0,30,4>0,3 →→ CC dư
Theo PT: nC(pư)nC(pư) == nCO2nCO2 == nO2nO2 == 0,3(mol)0,3(mol)
nC(dư)nC(dư) == 0,4−0,3=0,1(mol)0,4-0,3=0,1(mol)
mC(dư)mC(dư) == 0,1.12=1,2(g)0,1.12=1,2(g)
VCO2VCO2 == 0,3.22,4=6,72(l)
