mC = 1000.95% = 950 (g)
PT: \(C+O_2\underrightarrow{t^o}CO_2\)
Ta có: \(n_C=\dfrac{950}{12}=\dfrac{475}{6}\left(mol\right)\)
Theo PT: \(n_{O_2}=n_{CO_2}=n_C=\dfrac{475}{6}\left(mol\right)\)
a+b, \(V_{O_2}=V_{CO_2}=\dfrac{475}{6}.22,4\approx1773,33\left(l\right)\)