Em tham khảo nha:

\(\begin{array}{l}
1)\\
a)\\
{n_{C{O_2}}} = \dfrac{{1,32}}{{44}} = 0,03\,mol\\
{n_{{H_2}O}} = \dfrac{{0,54}}{{18}} = 0,03\,mol\\
{n_C} = {n_{C{O_2}}} = 0,03\,mol\\
{n_H} = 2{n_{{H_2}O}} = 0,06\,mol\\
{m_O} = 0,9 - 0,03 \times 12 - 0,06 = 0,48\,g\\
{n_O} = \dfrac{{0,48}}{{16}} = 0,03\,mol\\
{n_C}:{n_H}:{n_O} = 0,03:0,06:0,03 = 1:2:1\\
 \Rightarrow CTDGN:C{H_2}O\\
b)\\
CTPT\,X:{(C{H_2}O)_n}\\
{M_X} = 180g/mol \Rightarrow 30n = 180 \Rightarrow n = 6\\
CTPT:{C_6}{H_{12}}{O_6}\\
2)\\
{n_{C{O_2}}} = \dfrac{{0,792}}{{44}} = 0,018\,mol\\
{n_{{H_2}O}} = \dfrac{{0,297}}{{18}} = 0,0165\,mol\\
{n_C} = {n_{C{O_2}}} = 0,018\,mol\\
{n_H} = 2{n_{{H_2}O}} = 0,033\,mol\\
{m_O} = 0,513 - 0,018 \times 12 - 0,033 = 0,264\,g\\
{n_O} = \dfrac{{0,264}}{{16}} = 0,0165mol\\
{n_C}:{n_H}:{n_O} = 0,018:0,033:0,0165 = 12:22:11\\
 \Rightarrow CTDGN:{C_{12}}{H_{22}}{O_{11}}\\
CTPT\,X:{({C_{12}}{H_{22}}{O_{11}})_n}\\
{M_X} = 342g/mol \Rightarrow 342n = 342 \Rightarrow n = 1\\
CTPT:{C_{12}}{H_{22}}{O_{11}}\\
3)\\
{n_{C{O_2}}} = \dfrac{{13,44}}{{22,4}} = 0,6\,mol\\
{n_{{H_2}O}} = \dfrac{9}{{18}} = 0,5\,mol\\
{n_C} = {n_{C{O_2}}} = 0,6\,mol\\
{n_H} = 2{n_{{H_2}O}} = 1\,mol\\
{m_O} = 16,2 - 0,6 \times 12 - 1 = 8\,g\\
{n_O} = \dfrac{8}{{16}} = 0,5\,mol\\
{n_C}:{n_H}:{n_O} = 0,6:1:0,5 = 6:10:5\\
 \Rightarrow CTDGN:{C_6}{H_{10}}{O_5}\\
 \Rightarrow CTPT:{({C_6}{H_{10}}{O_5})_n} \Rightarrow X \text{ là tinh bột }
\end{array}\)

Câu 1:

a,

$n_C=n_{CO_2}=\dfrac{1,32}{44}=0,03(mol)$

$n_H=2n_{H_2O}=\dfrac{0,54}{18}.2=0,06(mol)$

$\to n_O=\dfrac{0,9-0,03.12-0,06.1}{16}=0,03(mol)$

$n_C: n_H: n_O=0,03:0,06:0,03=1:2:1$

$\to$ CTĐGN: $CH_2O$

b,

Đặt CTTQ: $(CH_2O)_n$

$\to (12+1.2+16)n=180$

$\to n=6$

Vậy cacbohidrat là $C_6H_{12}O_6$

Câu 2:

$n_C=n_{CO_2}=\dfrac{0,792}{44}=0,018(mol)$

$n_H=2n_{H_2O}=\dfrac{0,297}{18}.2=0,033(mol)$

$\to n_O=\dfrac{0,513-0,018.12-0,033.1}{16}=0,0165(mol)$

$n_C: n_H: n_O=0,018:0,033:0,0165=12:22:11$

$\to$ CTĐGN: $C_{12}H_{22}O_{11}$

Đặt CTTQ: $(C_{12}H_{22}O_{11})_n$

$\to (12.12+1.22+11.16)n=342$

$\to n=1$

Vậy cacbohidrat là $C_{12}H_{22}O_{11}$

Câu 3:

$n_C=n_{CO_2}=\dfrac{13,44}{22,4}=0,6(mol)$

$n_H=2n_{H_2O}=\dfrac{9}{18}.2=1(mol)$

$\to n_O=\dfrac{16,2-0,6.12-1.1}{16}=0,5(mol)$

$n_C: n_H: n_O=0,6:1:0,5=6:10:5$

$\to$ CTĐGN: $C_6H_{10}O_5$ 

$X$ là tinh bột hoặc xenlulozơ