Question

Bài 3 giúp mình với

Answer 1

\(a) CH_4+ 2O_2 \xrightarrow{t^o} CO_2 + 2H_2O\\ C_2H_4 + 3O_2 \xrightarrow{t^o} 2CO_2 + 2H_2O\\ b) n_{CH_4} = a(mol) ; n_{C_2H_4} = b(mol)\\ \Rightarrow a + b = \dfrac{4,48}{22,4} = 0,2(1)\\ n_{CO_2} = a + 2b = n_{CaCO_3} = \dfrac{25}{100} = 0,25(2)\\ (1)(2) \Rightarrow a = 0,15; b = 0,05\\ \%V_{CH_4} = \dfrac{0,15}{0,2}.100\% = 75\%\\ \%V_{C_2H_4} = 100\% -75\% = 25\%\\ c) n_{H_2O} = 2n_{CH_4} + 2n_{C_2H_4} = 0,4(mol) \Rightarrow a = 0,4.18 = 7,2(gam)\)