Em tham khảo nha :

\(\begin{array}{l}
1)\\
CTHH:{C_x}{H_y}{O_z} + (x + \dfrac{y}{4} - \dfrac{z}{2}){O_2} \to xC{O_2} + \dfrac{y}{2}{H_2}O\\
{n_{C{O_2}}} = \dfrac{m}{M} = \dfrac{{1,08}}{{44}} = 0,04mol\\
{n_{{H_2}O}} = \dfrac{m}{M} = \dfrac{{1,08}}{{18}} = 0,06mol\\
\text{Theo định luật bảo toàn khối lượng ta có :}\\
{m_{{O_2}}} + {m_A} = {m_{C{O_2}}} + {m_{{H_2}O}}\\
 \Rightarrow {m_{{O_2}}} = 1,08 + 1,76 - 1,24 = 1,6g\\
{n_{{O_2}}} = \dfrac{m}{M} = \dfrac{{1,6}}{{32}} = 0,05mol\\
{n_C} = {n_{C{O_2}}} = 0,04mol\\
{n_H} = 2{n_{{H_2}}} = 0,12mol\\
{n_O} = 2{n_{C{O_2}}} + {n_{{H_2}O}} - 2{n_{{O_2}}} = 0,04mol\\
x:y:z = 0,04:0,12:0,04 = 1:3:1\\
CTHH:{(C{H_3}O)_n}\\
{M_A} = 62dvC\\
 \Rightarrow (12 + 3 + 16)n = 62\\
 \Rightarrow n = \dfrac{{62}}{{31}} = 2\\
 \Rightarrow CTHH:{C_2}{H_6}{O_2}\\
2)\\
{m_{dv}} = 1000000g\\
{m_{CaC{O_3}}} = \dfrac{{1000000 \times 80}}{{100}} = 800000g\\
{n_{CaC{O_3}}} = \dfrac{m}{M} = \dfrac{{800000}}{{100}} = 8000mol\\
CaC{O_3} \to CaO + C{O_2}\\
{n_{CaO}} = {n_{CaC{O_3}}} = 8000mol\\
{m_{CaO}} = n \times M = 8000 \times 56 = 448000g\\
\text{Khối lượng CaO thu được là :}\\
{m_{CaO}} = \dfrac{{448000 \times 90}}{{100}} = 403200g
\end{array}\)