\(n_C=\dfrac{4,2}{12}=0,35\left(mol\right);n_{O_2}=\dfrac{10,6}{32}=0,33125\left(mol\right)\)
PTHH.C + O2 \(\underrightarrow{t^o}\) CO2
Mol: 0,33125 0,33125
Ta có:\(\dfrac{0,35}{1}>\dfrac{0,33125}{1}\) => C dư,O2 pứ hết
=> \(m_{CO_2}=0,33125.44=14,575\left(g\right)\)