\(n_{C_4H_{10}}=\dfrac{11.6}{58}=0.2\left(mol\right)\)
\(C_4H_{10}+\dfrac{13}{2}O_2\underrightarrow{^{t^0}}4CO_2+5H_2O\)
\(0.2.............1.3\)
\(n_{kk}=5n_{O_2}=1.3\cdot5=6.5\left(mol\right)\)
\(V_{kk}=6.5\cdot22.4=145.6\left(l\right)=0.1456\left(m^3\right)\)