\(n_{C_4H_{10}}=\frac{0,58}{58}=0,01\left(mol\right)\)
PTHH: \(2C_4H_{10}+13O_2-->8CO_2+10H_2O\) (câu a)
0,01 ----> 0,065--------> 0,02 ----> 0,05 (mol)
=>\(\left\{{}\begin{matrix}V_{O_2}=0,065.22,4=1,456\left(l\right)\\V_{CO_2}=0,02.22,4=0,448\left(l\right)\end{matrix}\right.\) (câu b)
=>\(\left\{{}\begin{matrix}m_{CO_2}=0,02.44=0,88\left(g\right)\\m_{H_2O}=0,05.18=0,9\left(g\right)\end{matrix}\right.\)(câu c)