\(n_C=\dfrac{m_C}{M_C}=\dfrac{4}{12}=\dfrac{1}{3}mol\)
\(C+O_2\rightarrow\left(t^o\right)CO_2\)
1/3 1/3 1/3 ( mol )
\(V_{O_2}=n_{O_2}.22,4=\dfrac{1}{3}.22,4=7,466l\)
\(m_{CO_2}=n_{CO_2}.M_{CO_2}=\dfrac{1}{3}.44=14,66g\)
nC = 4/12 = 1/3 (mol)
PTHH: C + O2 -> (t°) CO2
Mol: 1/3 ---> 1/3 ---> 1/3
VO2 = 1/3 . 22,4 = 22,4/3 (l)
mCO2 = 1/3 . 44 = 44/3 (g)
\(pthh:C+O_2\overset{t^o}{--->}CO_2\)
a. Ta có: \(n_C=\dfrac{4}{12}=\dfrac{1}{3}\left(mol\right)\)
Theo pt: \(n_{O_2}=n_C=\dfrac{1}{3}\left(mol\right)\)
\(\Rightarrow V_{O_2}=\dfrac{1}{3}.22,4\approx7,47\left(lít\right)\)
b. Theo pt: \(n_{CO_2}=n_C=\dfrac{1}{3}\left(mol\right)\)
\(\Rightarrow m_{CO_2}=\dfrac{1}{3}.44\approx14,7\left(g\right)\)
C+O2-to>CO2
\(\dfrac{1}{3}\)---\(\dfrac{1}{3}\)--------\(\dfrac{1}{3}\)
n C=\(\dfrac{4}{12}\)=\(\dfrac{1}{3}\) mol
=>VO2=\(\dfrac{1}{3}\).22,4=7,467l
=>m CO2=\(\dfrac{1}{3}\).44=14,67g
