a)
\(CO_2+NaOH\rightarrow Na_2CO_3+H_2O\)
\(2CO+O_2\rightarrow2CO_2\)
b)
\(n_{O2}=\frac{4,48}{22,4}=0,2\left(mol\right)\)
\(\rightarrow n_{CO}=0,2.2=0,4\left(mol\right)\)
\(V_{CO}=0,4.22,4=8,96\left(l\right)\)
\(V_{CO2}=16-8,96=7,04\left(l\right)\)
a) PTHH
CO2+NaOH-->Na2CO3+H2O
2CO+O2-->2CO2
b)nO2=4,4822,4=0,2(mol)
→nCO=0,2.2=0,4(mol
=>VCO=0,4.22,4=8,96(l)
=>VCO2=16−8,96=7,04(l)