\(C+O_2\underrightarrow{t^o}CO_2\)
\(n_{CO_2}=\dfrac{m}{M}=\dfrac{0,1}{44}=\dfrac{1}{440}\left(mol\right)\)
Theo PTHH, \(n_C=n_{CO_2}=\dfrac{1}{440}\left(mol\right)\)
\(m_C=n\cdot M=\dfrac{1}{440}\cdot12=\dfrac{3}{110}\left(g\right)\)
\(\%C\) trong thép là: \(\%C=\dfrac{m_C}{m_{thep}}=\dfrac{\dfrac{3}{110}}{5}\approx0,54\%\)