Đáp án:
$m=1,48\ (gam)$
Giải thích các bước giải:
$n_{CO_2}=\dfrac{4,4}{44}=0,1\ (mol)$
$\Rightarrow n_C=0,1\ (mol)$
$\Rightarrow m_C=0,1.12=1,2\ (gam)$
$n_{H_2O}=\dfrac{2,52}{18}=0,14\ (mol)$
$\Rightarrow n_H=0,14.2=0,28\ (mol)$
$\Rightarrow m_H=0,28.1=0,28\ (gam)$
$CH_4+2O_2\xrightarrow{t^\circ} CO_2+2H_2O$
$C_3H_8+5O_2\xrightarrow{t^\circ} 3CO_2+4H_2O$
$2C_4H_{10}+13O_2\xrightarrow{t^\circ} 8CO_2+10H_2O$
$\Rightarrow m=m_C+m_H=1,2+0,28=1,48\ (gam)$