a) \(\begin{array}{l}{n_{{O_2}}} = \dfrac{{4,48}}{{22,4}} = 0,2(mol)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C\,\,\,\,\,\, + \,\,\,\,\,{O_2} \to \,\,\,\,\,\,C{O_2}\\pt(mol)\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\\db(mol)\,\,0,3\,\,\,\,\,\,\,\,\,0,2\\Nx:\dfrac{{0,3}}{1} > \dfrac{{0,2}}{1}\end{array}\)
Cacbon dư sau phản ứng,
\(\begin{array}{l}{n_{C{O_2}}} = {n_{{O_2}}} = 0,2\,\,(mol)\\{m_{C{O_2}}} = 0,2 \times 44 = 8,8\,\,(gam)\end{array}\)
b) Làm như hướng dẫn giải ở phần a : \({m_{C{O_2}}} = 22(g)\).
