Em tham khảo nha :

\(\begin{array}{l}
a)\\
{n_{C{O_2}}} = \dfrac{{4,48}}{{22,4}} = 0,2mol\\
{n_C} = {n_{C{O_2}}} = 0,2mol\\
{n_{{H_2}O}} = \dfrac{{3,6}}{{18}} = 0,2mol\\
{n_H} = 2{n_{{H_2}O}} = 0,4mol\\
BT\,O:\\
{m_O} = 6 - 0,2 \times 12 - 0,4 = 3,2g\\
{n_O} = \dfrac{{3,2}}{{16}} = 0,2mol\\
CTPT:{C_x}{H_y}{O_z}\\
{n_A} = \dfrac{6}{{60}} = 0,1mol\\
x = \dfrac{{0,2}}{{0,1}} = 2\\
y = \dfrac{{0,4}}{{0,1}} = 4\\
z = \dfrac{{0,2}}{{0,1}} = 2\\
CTPT:{C_2}{H_4}{O_2}\\
b)\\
C{O_2} + Ca{(OH)_2} \to CaC{O_3} + {H_2}O\\
{n_{CaC{O_3}}} = {n_{C{O_2}}} = 0,2mol\\
{m_{CaC{O_3}}} = 0,2 \times 100 = 20g
\end{array}\)

$a,$

$n_{CO_2}=\dfrac{4,48}{22,4}=0,2(mol)$

$→n_{C}(A)=0,2(mol)$

$→m_{C}(A)=0,2×12=2,4(g)$

$n_{H_2O}=\dfrac{3,6}{18}=0,2(mol)$

$→n_{H}(A)=0,2×2=0,4(mol)$

$→m_{H}(A)=0,4(g)$

Ta có: $m_{C}+m_{H}=2,4+0,4=2,8(g)<m_{A}=6(g)$

$⇒$ A có nguyên tố $C,H,O$

Đặt CT A là $C_xH_yO_z$

$m_{O}(A)=6-2,8=3,2(g)$

$x:y:z=0,2:0,4:\dfrac{3,2}{16}=1:2:1$

$→CTĐG(A):\ (CH_2O)_n$

Ta có: $30n=60→n=2$

$→CT(A):\ C_2H_4O_2$

$b,$

$CO_2+Ca(OH)_2→CaCO_3+H_2O$

$\xrightarrow{BTNT\ C}n_{CaCO_3}=n_{CO_2}=0,2(mol)$

$→m_{CaCO_3}=0,2×100=20(g)$