a. \(CH_4+2O_2\rightarrow CO_2+2H_2O\)
b. \(V_{CH4}=1,5.97\%=1,455m^3=1455\left(l\right)\)
\(\rightarrow n_{CH2}=\frac{1455}{22,4}=64,95\left(mol\right)\)
\(n_{O2}=2n_{CH4}=2.64,95=129,9\left(mol\right)\)
\(\rightarrow V_{O2}=129,9.22,4=2909,76\left(l\right)=2,90976\left(m^3\right)\)
\(\rightarrow V_{kk}=\frac{2,90976}{20\%}=14,5488\left(m^3\right)\)